IP 

ACADEMIC  ALGEBRA-, 


FOR  THE  USE  OF 


COMMON   AND   HIGH   SCHOOLS   AND 
ACADEMIES. 


WITH  NUMEROUS  EXAMPLES. 


EDWARD  A.  BOWSER,  LL.D., 

// 

PROFESSOR  OF   MATHEMATICS   AND    ENGINEERING    IN   RUTGERS    COLLEGE. 


Oakland  High  School, 


BOSTOX,    U.S.A.:    ■ 
D.   C.    HEATH   &   CO.,   PUBLISHERS. 
1895. 


Copyright,  1888, 
By  E.  A.  BOWSEK 


Xorfajooti  $)rrss : 
Berwick  &  Smith,  Boston,  U.S.A. 


PREFACE. 


This  work  is  designed  as  a  text-book  for  Common  and 
High  Schools  and  Academies,  and  to  prepare  students  for 
entering  Colleges  and  Scientific  Schools.  The  book  is  a 
complete  treatise  on  Algebra  up  to  and  through  the  Pro- 
gressions, and  including  Permutations  and  Combinations 
and  the  Binomial  Theorem. 

The  aim  has  been  to  explain  the  principles  concisely  and 
clearly,  bestowing  great  care  upon  the  explanations  and 
proofs  of  the  fundamental  operations  and  rules.  Copious 
illustrations  have  been  given  to  make  the  work  intelligible 
and  interesting  to  young  students  ;  and  numerous  explan- 
atory notes  have  been  all  along  inserted,  to  guard  the  pupil 
against  the  errors  which  experience  shows  to  be  almost 
universal  among  beginners.  Thoroughness  has  been  aimed 
at,  rather  than  multiplicity  of  subjects.  If  a  student  has 
not  time  to  master  a  complete  course,  it  is  better  for  him 
to  omit  entirely  subjects  that  are  less  necessary,  than  to 
go  rapidly  over  too  many  things. 

In  the  earlier  chapters,  some  of  the  most  interesting 
practical  applications  of  the  subject  have  been  introduced. 
Thus,  a  chapter  on  easy  equations  and  problems  precedes 
the  chapters  on  Factoring  and  Fractions.  By  this  course 
the  beginner  soon  becomes  acquainted  with  the  ordinary 
Algebraic  processes  without  encountering  too  many  of  their 
difficulties  ;  and  he  is  learning  at  the  same  time  something 
of  the  more  attractive  parts  of  the  subject.      Nothing  is 

A  Q  o  n  *  r^ 


VI  PREFACE. 

more  pleasing  to  a  young   student  than   to   see   and   feel 
that  he  can  use  his  knowledge  to  some  practical  end. 

Throughout  the  book  are  numerous  examples  fully  worked 
out,  to  illustrate  the  most  useful  applications  of  important 
rules,  and  to  exhibit  the  best  methods  of  arranging  the  work. 
No  principle  is  well  learned  by  a  pupil  and  thoroughly  fixed 
in  his  mind  till  he  can  use  it.  For  this  purpose  a  large 
number  of  examples  is  given  at  the  ends  of  the  chapters. 
These  examples  have  been  selected  and  arranged  so  as 
to  illustrate  and  enforce  every  part  of  the  subject.  Each 
set  has  been  carefully  graded,  commencing  with  some  which 
are  very  easy,  and  proceeding  to  others  which  are  more 
difficult.  Complicated  examples  have  been  excluded, 
because  they  consume  time  and  energy  which  may  be 
spent  more  profitably  on  other  branches  of  mathematics. 

The  chief  sources  from  which  I  have  derived  assistance 
in  preparing  this  work  are  the  treatises  of  Wood,  De 
Morgan,  Serret,  Todhunter,  Colenzo,  Hall  and  Knight, 
Smith,  and  Chrystal. 

My  thanks  are  due  to  those  of  my  friends  who  have 
kindly  assisted  me  in  reading  the  MS.,  correcting  the 
proof-sheets,  and  verifying  copy. 

E.    A.    B. 

Rutoers  College, 
New  Brunswick,  N.J.,  June,  1888. 


TABLE    OF    CONTENTS. 


CHAPTER  I. 

FIRST     PRINCIPLES. 

ART.  PAGE 

1.  Quantity  and  its  Measure 1 

2.  Number 1 

3.  Mathematics 2 

4.  Algebra 2 

5.  Algebraic  Symbols 2 

6.  Symbols  of  Quantity 2 

7.  Symbols  of  Operation 3 

S.  The  Sign  of  Addition 3 

9.  The  Sign  of  Subtraction 3 

10.  The  Sign  of  Multiplication 4 

11.  The  Sign  of  Division 5 

12.  The  Exponential  Sign 6 

13.  The  Radical  Sign 7 

14.  Symbols  of  Relation 7 

15.  Symbols  of  Abbreviation 8 

16.  Algebraic  Expressions 9 

17.  Factor  — Coefficient 11 

18.  A  Term,  its  Dimensions,  and  Degree  —  Homogeneous  ...  12 

19.  Simple  and  Compound  Expressions 13 

20.  Positive  and  Negative  Quantities 13 

21.  Additions  and  Multiplications  Made  in  any  Order    ....  17 

22.  Suggestions  for  the  Student  in  Solving  Examples     ....  IS 
Examples 18 

CHAPTER   II. 

ADDITION. 

23.  Addition  —  Algebraic  Sum 20 

24.  To  Add  Terms  which  are  Like  and  have  Like  Signs     ...  20 

25.  To  Add  Terms  which  are  Like,  but  have  Unlike  Signs     .     .  21 

26.  To  Add  Terms  which  are  not  all  Like  Terms 21 

27.  Remarks  on  Addition 22 

Examples 23 

vii 


Vlll  CONTENTS. 


CHAPTER  III. 

SUBTRACTION. 

k-RT.  PAGE 

28.  Subtraction  —  Algebraic  Difference 25 

29.  Rule  for  Algebraic  Subtraction 25 

30.  Remarks  on  Addition  and  Subtraction 28 

31.  The  Use  of  Parentheses 29 

32.  Plus  Sign  before  the  Parenthesis 30 

33.  Minus  Sign  before  the  Parenthesis 30 

34.  Compound  Parentheses 31 

Examples 32 


CHAPTER  IV. 

MULTIPLICATION. 

35.  Multiplication 36 

36.  Rule  of  Signs 36 

37.  The  Multiplication  of  Monomials 39 

38.  To  Multiply  a  Polynomial  by  a  Monomial 40 

39.  Multiplication  of  a  Polynomial  by  a  Polynomial 41 

40.  Multiplication  by  Inspection 46 

41.  Special  Forms  of  Multiplication  —  Formulas 47 

42.  Important  Results  in  Multiplication 51 

43.  Results  of  Multiplying  Algebraic  Expressions 52 

Examples 53 


/ 


CHAPTER  V. 

DIVISION. 

44.  Division 55 

45.  The  Division  of  one  Monomial  by  Another 55 

46.  The  Rule  of  Signs 57 

47.  To  Divide  a  Polynomial  by  a  Monomial 58 

48.  To  Divide  one  Polynomial  by  Another 58 

49.  Division  with  the  Aid  of  Parentheses 64 

50.  Where  the  Division  cannot  be  Exuetly  Performed     ....  64 

51.  Important  Examples  in  Division 65 

Examples 6(5 


CONTENTS.  ix 


CHAPTER   VI. 

SIMPLE    EQUATIONS    OF    ONE    UNKNOWN    QUANTITY. 
ART.  PAGE 


52.  Equations  —  Identical  Equations 68 

53.  Equation  of  Condition  —  Unknown  Quantity 68 

54.  Axioms 69 

Clearing  of  Fractions    ....         70 


OO 


56.  Transposition 71 

57.  Solution  of  Simple  Equations  with  one  Unknown  Quantity  .    72 


58.  Fractional  Equations 75 

59.  To  Solve  Equations  whose  Coefficients  are  Decimals    ...  77 

60.  Literal  Equations 77 

61.  Problems  Leading  to  Simple  Equations 79 

Examples 80 


CHAPTER  VII. 

FACTORING  —  GREATEST   COMMON    DIVISOR LEAST   COMMON 

MULTIPLE. 

62.  Definitions 90 

63.  When  All  the  Terms  Have  one  Common  Factor 90 

64.  Expressions  Containing  Four  Terms 91 

65.  To  Factor  a  Trinomial  of  the  Form  x'2  +  ax  +  6     ....  93 

66.  To  Factor  a  Trinomial  of  the  Form  ax2  +  bx  +  c    .     .     .     .  95 

67.  To  Factor  the  Difference  of  Two  Squares 98 

68.  When  One  or  Both  of  the  Squares  is  a  Compound  Expression  99 

69.  Compound  Quantities  as  the  Difference  of  Two  Squares  .    .  100 

70.  To  Factor  the  Sum  or  the  Difference  of  Two  Cubes      .     .     .101 

71.  Miscellaneous  Cases  of  Resolution  into  Factors 101 

Examples 102 

GREATEST   COMMON    DIVISOR. 

72.  Definitions 105 

73.  Monomials,  and  Polynomials  which  ran  be  Easily  Factored  .  105 

74.  Expressions  which  cannot  be  Easily  Factored 1<>7 

Examples 109 


CONTENTS. 


LEAST    COMMON   MULTIPLE. 

ART.  PAGE 

75.  Definitions 114 

7G.  Monomials,  and  Polynomials  which  can  be  Easily  Factored  .  115 

77.  Expressions  which  cannot  be  Easily  Factored 117 

Examples Ill' 


CHAPTER   VIII. 

FRACTIONS. 

78.  A  Fraction  —  Entire  and  Mixed  Quantities .     . 121 

79.  To  Reduce  a  Fraction  to  its  Lowest  Terms 122 

80.  To  Reduce  a  Mixed  Quantity  to  the  Form  of  a  Fraction    .     .  125 

81.  To  Reduce  a  Fraction  to  an  Entire  or  Mixed  Quantity .     .     .126 

82.  To  Reduce  Fractions  to  their  Least  Common  Denominator  .  127 

83.  Rule  of  Signs  in  Fractions 128 

84.  Addition  and  Subtraction  of  Fractions 131 

85.  To  Multiply  a  Fraction  by  an  Integer      ........  137 

36.  »To  Divide  a  Fraction  by  an  Integer 137 

37.  To  Multiply  Fractions 138 

88.  To  Divide  Fractions 140 

89.  Complex  Fractions 141 

90.  A  Single  Fraction  Expressed  as  a  Group  of  Fractions  .     .     .  144 
Examples 145 

CHAPTER  \£t. 

HARDER    SIMPLE    EQUATIONS    OF    ONE    UNKNOWN    QUANTITY. 

91.  Solution  of  Harder  Equations    .     .     .  *  „ 149 

92.  Harder  Problems  Leading  to  Simple  Equations 153 

Examples 162 


&R 


CHAPT 

SIMULTANEOUS   SIMPLE   EQUATIONS   OF  TWO    OR   MORE    UNKNOWN 
QUANTITIES. 

93.  Simultaneous  Equations  of  Two  Unknown  Quantities      .     .  169 

94.  Elimination 171 

95.  Elimination  by  Addition  or  Subtraction 171 


CONTENTS.  xi 


ART.  TAfJE 

06.  Elimination  by  Substitution 175 

07.  Elimination  by  Comparison 170 

08.  Fractional  Simultaneous  Equations ITS 

00.  Literal  Simultaneous  Equations 171) 

100.  Simultaneous  Equations  with  Three  Unknown  Quantities    .  ISO 

101.  Problems  Leading  to  Simultaneous  Equations 185 

Examples 101 


CHAPTER  XI. 

INVOLUTION    ANJ>   EVOLUTION. 

102.  Involution 108 

103.  Involution  of  Powers  of  Monomials *    .     .  108 

104.  Involution  of  Binomials 200 

.105.  Involution  of  Polynomials 201 

EVOLUTION. 

106.  Evolution  —  Evolution  of  Monomials 203 

107.  Square  Koot  of  a  Polynomial 205 

108.  Square  Root  of  Arithmetic  Numbers 200 

100.  Square  Root  of  a  Decimal 211 

110.  Cube  Root  of  a  Polynomial 214 

111.  Cube  Root  of  Arithmetic  Numbers 217 

112.  Cube  Root  of  a  Decimal 210 

Examples     ...,,,,,, 220 


CHAPTER   XII. 

THE  THEORY  OF  EXPONENTS  SURDS. 

113.  Exponents  that  are  Positive  Integers 223 

114.  Fractional  Exponents 223 

115.  Negative  Exponents 225 

116.  To  Prove  that  (a")»  =  amn  is  True  for  All  Values  of  m  and  n    227 

117.  To  Prove  that  [ab)n  =  anbn  for  Any  Value  of  n 228 

SURDS     (RADICALS). 

118.  Surds  — Definitions 231 

110.  To  Reduce  a  Rational  Quantity  to  a  Surd  Form 238 

120.  To  Introduce  the  Coefficient  Under  the  Radical  Sign    .     .     .  232 


xii  CONTENTS. 

ART.  PAGE 

121.  To  Reduce  an  Entire  to  a  Mixed  Surd 232 

122.  Reduction  of  Surds  to  Equivalent  Surds 233 

123.  Addition  and  Subtraction  of  Surds 234 

124.  Multiplication  of  Surds 235 

125.  To  Rationalize  the  Denominator  of  a  Fraction 237 

120.  Division  of  Surds 238 

127.  Binomial  Surds  — Important  Propositions 239 

128.  Square  Root  of  a  Binomial  Surd .240 

129.  Equations  Involving  Surds 242 

Examples 243 


CHAPTER   XIII. 

QUADRATIC  EQUATIONS  OF  ONE  UNKNOWN  QUANTITY. 

130.  Quadratic  Equations 240 

131.  Pure  Quadratic  Equations 240- 

132.  Affected  Quadratic  Equations 251 

133.  Condition  for  Equal  Roots 255 

134.  Hindoo  Method  of  Completing  the  Square 257 

135.  Solving  a  Quadratic  by  Factoring 259 

130.  To  Form  a  Quadratic  when  the' Roots  are  Given 201 

137.  Equations  Having  Imaginary  Roots 2G3 

138.  Equations  of  Higher  Degree  than  the  Second 204 

139.  Solutions  by  Factoring 207 

140.  Problems  Leading  to  Quadratic  Equations 209 

Examples 272 

CHAPTER   XIV. 

SIMULTANEOUS    QUADRATIC    EQUATIONS. 

141.  Simultaneous  Quadratic  Equations 279 

142.  When  One  of  the  Equations  is  of  the  First  Degree  ....  279 

143.  Equations  of  the  Form  x  ±  y  =  a,  and  xy  =  b 280 

144.  When  the  Equations  Contain  a  Common  Algebraic  Factor    .  282 

145.  Homogeneous  Equations  of  the  Second  Degree 2S4 

140.  When  the  Two  Equations  are  Symmetrical 280 

147.  Special  Methods 287 

148.  Quadratic  Equations  with  Three  Unknown  Quantities      .     .  289 

149.  Problems  Leading  to  Simultaneous  Quadratic  Equations  .     .  289 
Examples 292 


CONTENTS.  xiii 


CHAPTER  XV. 

RATIO  —  PROPORTION  —  VARIATION. 

ART.  PAGK 

150.  Ratio  —  Definitions 297 

151.  Properties  of  Ratios 299 

PROPORTION. 

152.  Definitions 302 

103.  Properties  of  Proportions 303 

VARIATION. 

154.  Definition 308 

155.  Different  Cases  of  Variation 301) 

156.  Propositions  in  Variation .  310 

Examples 314 


CHAPTER    XVI. 

ARITHMETIC,    GEOMETRIC,    AND    HARMONIC    PROGRESSIONS. 

ARITHMETIC    PROGRESSION. 

157.  Definitions  — Formulae 318 

158.  Arithmetic  Mean 321 

GEOMETRIC    PROGRESSION. 

159.  Definition  — Formula? 324 

160.  Geometric  Mean 326 

161.  The  Sum  of  an  Infinite  Number  of  Terms 327 

162.  Value  of  a  Repeating  Decimal 329 

HARMONIC    PROGRESSION. 

163.  Definition 330 

164.  Harmonic  Mean 331 

165.  Relation  between  the  Different  Means     „ 333 

Examples 334 


xiv  CONTENTS. 

CHAPTER   XVII. 

PERMUTATIONS    AND    COMBINATIONS  BINOMIAL   THEOREM. 


PERMUTATIONS    AND    COMBINATIONS. 

ART.  PAGE 

166.  Definitions  . 338 

167.  The  Number  of  Permutations 339 

168.  The  Number  of  Combinations 341 

169.  To  Divide  m  +  n  Things  into  Two  Classes 343 

170.  Permutations  of  n  Things  not  all  Different 344 

BINOMIAL   THEOREM. 

171.  Positive  Integral  Exponent 345 

172.  The  rth  or  General  Term  of  the  Expansion 348 

Examples 350 


ALGEBRA. 


CHAPTER    I. 

FIRST     PRINCIPLES. 

1.  Quantity  and  its  Measure.  —  Quantity  is  any 
thing  that  is  capable  of  increase,  diminution,  and  measure- 
ment;  as  time,  space,  motion,  weight,  and  area. 

To  measure  a  quantity  is  to  find  how  many  times  it  con~ 
tains  another  quantity  of  the  same  kind,  taken  as  a  standard 
of  comparison.     This  standard  is  called  a  unit. 

For  example,  if  we  wish  to  determine  the  quantity  of  a  weight,  we 
must  take  a  unit  of  weight,  such  as  a  pound,  or  an  ounce,  and  observe 
how  many  times  it  is  contained  in  the  quantity  to  be  measured.  If 
we  wish  to  measure  area,  we  must  take  a  unit  of  area,  as  a  square 
foot,  square  yard,  or  acre,  and  see  how  many  times  it  is  contained  in 
the  area  to  be  measured.  So  also,  if  we  wish  to  measure  the  value  of 
a  sum  of  money,  or  any  portion  of  time,  we  must  take  a  unit  of  value, 
as  a  dollar  or  a  sovereign,  or  a  unit  of  time,  as  a  day  or  a  year,  and  see 
how  many  times  it  is  contained  in  the  quantity  to  be  measured. 

2.  Number.  —  The  relation  between  any  quantity  and  its 
unit  is  always  expressed  by  a  number;  a  number  therefore 
simply  shows  how  many  times  an}'  quantity  to  be  measured 
contains  another  quantity,  arbitrarily  assumed  as  the  unit. 
All  quantities,  therefore,  can  be  expressed  by  numbers. 

All  numbers  are  concrete  or  abstract. 

A  Concrete  Number  is  one  in  which  the  kind  of  quantity 
which  it  measures  is  expressed  or  understood ;  as  G  books, 
10  men,  4  days. 

1 


2  MAT  II  EM  A  TICS.  —  ALGEBRA. 

An  Abstract  Number  is  one  in  which  the  kind  of  quantity 
which  it  measures  is  not  expressed  ;  as  6,  10,  4. 

The  word  quantity  is  often  used  with  the  same  meaning  as  number. 
Numbers  may  be  either  whole  or  fractional.  The  word  integer  is  often 
used  instead  of  whole  number. 

3.  Mathematics.  —  Mathematics  is  the  science  which 
treats  of  the  measurement  and  relations  of  quantities.  It 
is  divided  into  two  parts,  Pure  Mathematics  and  Mixed 
Mathematics. 

Pure  Mathematics  consists  of  the  four  branches,  Arithmetic, 
Algebra,  Geometry,  and  Calculus. 

.  Mixed  Mathematics  is  the  application  of  Pure  Mathematics 
to  the  Mechanic  Arts. 

4.  Algebra.  —  Algebra  is  that  branch  of  Mathematics  in 
which  we  reason  about  numbers  by  means  of  symbols.  The 
different  symbols  used  represent  the  numbers  themselves, 
the  manner  in  which  they  are  related  to  one  another,  and  the 
operations  performed  on  them. 

In  Arithmetic,  numbers  are  represented  by  ten  characters,  called 
figures,  which  are  variously  combined  according  to  certain  rules,  and 
which  have  but  one  single  definite  value.  In  Algebra,  on  the  contrary, 
numbers  are  represented  either  by  figures,  as  in  Arithmetic,  or  by 
symbols  which  may  have  any  value  we  choose  to  assign  to  them. 

5.  Algebraic  Symbols.  —  The  symbols  employed  in 
Algebra  are  of  four  kinds :  symbols  of  quantity,  symbols  of 
operation,  symbols  of  relation,  and  symbols  of  abbreviation. 

6.  Symbols  of  Quantity.  —  The  symbols  of  quantity 
may  be  any  characters  whatever,  but  those  that  are  most 
commonly  used  are  figures  and  the  letters  of  the  alphabet; 
and  as  in  the  simplest  mathematical  problems  there  are 
certain  quantities  given,  in  order  to  determine  other  quan- 
tities which  are  unknown,  it  is  usual  to  represent  the  known 
quantities  by  figures  and  by  the  first  letters  of  the  alphabet, 
a,  b,  c,  etc.  ;  a',  V ',  c' ,  etc.,  rend  a  prime,  b  prime,  c  prime, 
etc.  ;  tiv  bv  cv  etc.,  read  a  one,  b  one,  c  one,  etc.  ;  while  the 


SYMBOLS   OF   OPERATION.  3 

unknown  quantities  are  represented  by  the  final  letters  of 
the  alphabet,  v,  x,  y,  z,  v',  x',  y',  z',  etc. 

Known  Quantities  are  those  whose  values  are  given. 

Unknoicu  Quantities  are  those  whose  values  are  required. 

Since  all  quantities  can  be  expressed  by  numbers  (Art.  2), 
it  is  only  these  numbers  with  which  we  are  concerned,  and 
the  symbols  of  quantity,  whether  figures  or  letters,  always 
represent  numbers. 

In  Arithmetic  a  character  has  but  one  definite  and  invariable  value, 
while  in  Algebra  a  symbol  may  stand  for  any  quantity  we  choose  to 
assign  to  it  (Art.  4);  but  while  there  is  no  restriction  as  to  the 
numerical  values  a  symbol  may  represent,  it  is  understood  that  in  the 
same  piece  of  work  it  keeps  the  same  value  throughout.  Thus,  when 
we  say  "let  a  =  2,"  we  do  not  mean  that  a  must  have  the  value  2 
always,  but  only  in  the  particular  example  we  are  considering.  Also, 
we  may  operate  with  symbols  without  assigning  to  them  any  particular 
value  at  all;  and  it  is  with  such  operations  that  Algebra  is  chiefly 
concerned. 

7.  Symbols  of  Operation.  —  The  symbols  of  operation 
are  the  same  in  Algebra  as  in  Arithmetic,  or  in  any  other 
branch  of  Mathematics,  and  are  the  following : 

8.  The  Sign  of  Addition,  +,  is  called  plus.  When 
placed  before  a  number  it  denotes  that  the  number  is  to 
be  added.  Thus,  6  -f-  3,  read  6  plus  3,  means  that  3  is  to  be 
added  to  6  ;  a  +  b.  read  a  j^us  b,  denotes  that  the  number 
represented  by  b  is  to  be  added  to  the  number  represented 
by  a  ;  or,  more  briefly,  it  denotes  that  b  is  to  be  added  to  a. 
If  a  represent  8,  and  b  represent  5,  then  a  -\-  b  represents  13. 

Similarly  a  +  b  +  c,  read  a  p/?*s  b  irtus  c,  denotes  that 
we  are  to  add  b  to  a,  and  then  add  c  to  the  result. 

9.  The  Sign  of  Subtraction,  — ,  is  called  minus.  When 
placed  before  a  number  it  denotes  that  the  number  is  to  be 
subtracted.  Thus,  a  —  6,  read  a  minus  6,  denotes  that  the 
number  represented  by  b  is  to  be  subtracted  from  the  number 
represented  by  a  ;  or,  more  briefly,  that  b  is  to  be  subtracted 
from  a.  If  a  represent  8,  and  b  represent  5,  then  a  —  b 
represents  3. 


4  THE   SrGN    OF  MULTIPLICATION. 

Similarly  a  —  b  —  c,  read  a  minus  b  minus  c,  denotes 
that  we  are  to  subtract  b  from  a,  and  then  subtract  c  from 
the  result. 

If  neither  +  nor  —  stands  before  a  quantity,  +  is  always 
understood  ;  thus  a  means  +  «. 

Quantities  which  have  the  same  sign,  either  +  or  — ,  are 
said  to  have  like  signs.  Thus,  +  «  and  -f-  b  have  like  sigus, 
also  —  a  and  —  b ;  but  -+-  a  and  —  b  have  unlike  signs. 

Note.  —  Although  there  are  many  signs  used  in  Algebra,  when  the 
sign  of  a  quantity  is  spoken  of,  it  means  the  +  or  —  sign  which  is 
prefixed  to  it;  and  when  we  speak  of  changing  the  signs  of  an  expres- 
sion, it  means  that  we  are  to  change  +  to  —  and  —  to  +  wherever 
they  occur. 

The  sign  ~  is  sometimes  used  to  denote  the  difference  of  two 
numbers  when  it  is  not  known  which  of  them  is  the  greater.  Thus, 
a  ~  b  denotes  the  difference  of  the  numbers  represented  by  a  and  &; 
and  is  equal  to  a  —  b,  or  b  —  a,  according  as  a  is  greater  or  less  than  6; 
but  this  symbol  ~  is  very  rarely  required. 

10.  The  Sign  of  Multiplication,  x ,  is  read  into,  or 
times,  or  multiplied  by.  When  placed  between  two  numbers 
it  denotes  that  they  are  to  be  multiplied  together.  Thus, 
a  x  6,  read  a  into  6,  denotes  that  the  number  represented  by 
a  is  to  be  multiplied  by  the  number  represented  by  6,  or, 
more  briefly,  that  a  is  to  be  multiplied  by  6,  or  that  the  two 
are  to  be  multiplied  together.  The  numbers  to  be  multiplied 
together  are  called  factors,  and  the  result  of  the  multiplica- 
tion is  called  a  product.  Thus  5,  a,  and  b  are  the  factors 
of  the  product  5  x  a  X  b.  If  a  represent  8,  and  b  repre- 
sent 4,  then  a  x  b  represents  32  ;  a  and  b  are  the  factors  of 
the  product  a  X  b,  or  8  and  1  are  the  factors  of  o'2.  Simi- 
larly a  x  b  x  c  denotes  the  product  of  the  numbers  a,  b, 
and  c.  If  a  represent  G,  b  represent  8,  and  c  represent  10, 
then  a  x  b  X  c  represents  180,  and  5  x  a  x  b  x  c  repre- 
sents 2400. 

Sometimes  apoint  is  used  instead  of  the  sign  x  ;  or,  still 
more  commonly,  one  number  is  placed  close  after  the  oilier 


THE  SIGN   OF  DIVISION.  5 

without  any  sign  between  them.  Thus,  a  x  6,  a  •  6,  and  ab 
all  mean  the  same  thing,  viz.,  the  product  of  a  and  b ;  also, 
a  X  b  x  c,  or  a  •  b  •  c,  or  afrc,  denotes  the  product  of  the 
numbers  a,  5,  and  c.  If  a,  6,  and  c  represent  2,  5,  and  10 
respectively,  then  abc  represents  100. 

If  one  factor  of  a  product  is  equal  to  0,  the  whole  product 
must  be  equal  to  0,  whatever  values  the  other  factors  may 
have.     A  factor  0  is  sometimes  called  a  "zero  factor."  * 

The  sign  of  multiplication  must  not  be  omitted  when  num- 
bers are  expressed  in  the  ordinary  way  by  figures.  Thus  23 
cannot  be  used  to  represent  the  product  of  2  and  3,  because 
23  is  used  to  mean  the  number  twenty-three.  Nor  can  the 
product  of  2  and  3  be  represented  by  2.3,  because  2.3  is 
used  to  mean  tioo  and  three-tenths.  We  must  therefore 
represent  the  product  of  2  and  3  by  placing  the  sign  of 
multiplication  between  them,  as  follows:  2x3.  When  the 
numbers  to  be  multiplied  together  are  represented  by  letters, 
or  by  letters  and  a  figure,  it  is  usual  to  omit  the  sign  of 
multiplication  for  the  sake  of  brevity,  and  write  them  in 
succession  close  to  each  other ;  thus,  the  product  of  the 
numbers  7,  «,  b,  c,  and  cl  would  be  written  labcd,  instead  of 
7  x  a  x  b  x  c  x  cZ,  or  7  •  a  •  b  •  c  •  d,  and  would  have  the 
same  meaning. 

11.  The  Sign  of  Division,  -*-,  is  read  divided  by,  or 
simply  by.  When  placed  between  two  numbers,  it  denotes 
that  the  number  which  precedes  it  is  to  be  divided  by  the 
number  which  follows  it.  Thus,  a  -f-  b,  read  a  divided  by  6, 
or  a  by  6,  denotes  that  the  number  represented  by  a  is  to  be 
divided  by  the  number  represented  by  &,  or,  more  briefly, 
that  a  is  to  be  divided  by  b.  If  a  represent  8,  and  b  repre- 
sent 2,  then  a  -~  b  represents  4.  Most  frequentl}*,  to  express 
division,  the  number  to  be  divided  is  placed  over  the  other 


*  It  is  a  common  mistake  of  beginners  to  say  that  an  Algebraic  expression  like 
a  x  0  or  0  x  a  is  equal  to  a,  by  supposing  it  to  mean  a  not  multiplied  at  all ;  whereas 
(i  x  0  ■  r  0  x  u  signifies  0  taken  a  limes,  or  a  taken  0  times,  and  is  therefore  equal 
toO. 


THE   EXPONENTIAL   SIGN. 


with   a  horizontal  line  between  them,  in  the  manner  of   a 

fraction  in  Arithmetic.     Thus,  -  is  used  instead  of  a  -r-  b, 

b 

and  has  the  same  meaning.     Also,  the  sign  of  division  may 

be  replaced  by  a  vertical  line,  straight  or  curved.     Thus, 

a  [6,  or  bja  is  used  instead  of  a  -h  b,   and  has  the  same 


Note.  —  It  is  important  for  the  student  to  notice  the  order  of  the 
operations  in  such  expressions  as  a  +  b  x  c  and  a  —  b  +  c.  The 
former  means  that  b  is  first  to  be  multiplied  by  c,  and  the  result  added 
to  a.  The  latter  means  that  b  is  first  to  be  divided  by  c,  and  the  result 
subtracted  from  a. 

12.  The  Exponential  Sign.  —  This  sign  is  a  small  fig- 
ure or  letter  written  at  the  right  of  and  above  a  number  to 
show  how  many  times  the  number  is  taken  as  a  factor,  and 
is  called  an  exponent.  Thus,  a2  is  used  to  denote  a  x  o,  or 
that  a  is  taken  twice  as  a  factor ;  a3  is  used  to  denote  a  x  a 
X  a,  or  that  a  is  taken  three  times  as  a  factor ;  a4  is  used  to 
denote  a  X  a  X  a  X  a,  or  that  a  is  taken  four  times  as  a 
factor;  and  an  is  used  to  denote  a  x  a  x  a  x  a,  etc.,  to 
n  factors,  or  that  a  is  taken  n  times  as  a  factor.  Similarly 
a?b*cd*  is  used  to  denote  aabbbbcddd,  and  7a3cd2  is  used  for 
laaacdd. 

If  a  factor  be  multiplied  by  itself  any  number  of  times 
the  product  is  called  a  power  of  that  factor.     Thus, 

a  x  a  is  called  the  second  power  of  a,  and  is  written  a2; 
a  x  a  x  (Us  called  the  third  power  of  a,  and  is  written  «3; 
a  x  a  X  a  X  a  is  called  the  fourth  power  of  a,  and  is  written  a4; 

and  so  on.  Similarly  aaabbc  is  called  the  product  of  the 
third  power  of  a,  the  second  power  of  6,  and  c,  and  is  written 
a3b2c. 

The  second  power  of  a,  i.e.,  a2,  is  usually  read  a  to  the 
second  power,  or  a  square.  The  third  power  of  «,  i.e.,  «3, 
is  usually  read  a  to  the  third  power,  or  a  cube.  There  are 
no  such  words  in  use  for  the  higher  powers ;  the  fourth 
power  of  a,  i  e.,  a4,  is  usually  read  a  to  the  fourth  power, 


THE  RADICAL   SIGN —  SYMBOLS    OF  RELATION.  7 

or  briefly,  a  fourth  power  ;  and  so  on.     "When  the  exponent 

is  unity  it  is  omitted.     Thus  we  do  not  write  a1,  but  simply 
a,  which  is  the  same  as  a1,  and  means  a  to  the  first  power, 

13.  The  Radical  Sign,  V7  .  — A  root  of  a  quantity  is  a 
factor,  which,  multiplied  by  itself  a  certain  number  of  times, 
will  produce  the  given  quantity.  The  square  root  of  a 
quantity  is  that  quantity  whose  square  or  second  power  is 
equal  to  the  given  quantity.  Thus  the  square  root  of  16 
is  4,  because  42  is  equal  to  16  ;  the  square  root  of  a2  is  a,  of 
81  is  9. 

The  square  root  of  a  is  denoted  by  va,  or  more  simply 

Similarly  the  cube,  fourth,  fifth,  etc.,  root  of  any  quantity 
is  that  quantity  whose  third,  fourth,  fifth,  etc.,  power  is 
equal  to  the  given  quantity. 

The  roots  are  denoted  by  the  symbols  V  ,  V  ?  V  j  etc. ; 
thus,  y-21as  denotes  the  cube  root  of  27a3,  which  is  3a, 
because  3a  to  the  third  power  is  27a3.  Similarly  ^32  is  2. 
The  small  figure  placed  on  the  left  side  of  the  symbol  is 
called  the  index  of  the  root.  Thus  2  is  the  index  of  the 
square  root,  3  of  the  cube  root,  4  of  the  fourth  root,  and 
so  on  ;  the  index,  however,  is  generally  omitted  in  denoting 
the  square  root ;  thus  \a  is  written  instead  of  \/a. 

The  symbol  V7"  is  sometimes  called  the  radical  sign. 
"When  this  sign  with  the  proper  index  on  the  left  side  of  it 
is  placed  over  a  quantity  it  denotes  that  some  root  of  the 
quantity  is  to  be  extracted. 

14.  Symbols  of  Relation.  —  The  symbols  of  relation 
are  the  following : 

The  sign  of  equality,  =  ,  is  read  equals,  or  is  equal  to. 
When  placed  between  two  numbers,  it  denotes  that  they  are 
equal  to  each  other.  Thus  a  =  b,  read  a  equals  b,  or  a  is 
equal  to  b,  denotes  that  the  number  represented  by  a  is  equal 
to  the  number  represented  by  b\  or,  more  briefly,  that  a 
equals  b.     And   a  -}-  b  =  c  denotes    that   the    sum   of   the 


8  SYMBOLS   OF  ABBREVIATION. 

numbers  a  and  b  is  equal  to  the   number  c ;    so   that  if  a 
represent  8  and  b  represent  4,  then  c  must  represent  12. 

The  sigyis  of  inequality,  >  and  <,  are  read  is  greater  than, 
and  is  less  than,  respectively.  When  either  is  placed  between 
two  numbers  it  denotes  that  they  are  unequal  to  each 
other,  the  opening  of  the  angle  in  both  cases  being  turned 
towards  the  greater  number.  Thus  a  >  b,  read  a  is  greater 
than  b,  denotes  that  the  number  a  is  greater  than  the  number 
6,  and  b  <  a,  read  b  is  less  than  a,  denotes  that  the  number  b 
is  less  than  the  number  a. 

The  sign  of  ratio,  : ,  is  read  is  to  or  to.  When  placed 
between  two  numbers  it  denotes  their  ratio.  Thus  a  :  b, 
read  a  is  to  b,  or  the  ratio  of  a  to  b,  denotes  the  ratio  of  the 
number  a  to  the  number  b.  A  proportion,  or  two  equal 
ratios,  is  expressed  by  writing  the  sign  =  or  the  sign  :  : 
between  two  equal  ratios.     Thus 

a :  b  —  c  :  d,  or  a  :  b  :  :  c  :  d, 
read  a  is  to  b  as  c  is  to  d,  or  the  ratio  of  a  to  b  equals  the 
ratio  of  c  to  d. 

The  sign  of  variation,  <x  ,  is  read  varies  as.  When  placed 
between  two  numbers  it  denotes  that  they  increase  and 
decrease  together,  in  the  same  ratio.  Thus  x  a  y,  read  x 
varies  as  y,  denotes  that  x  and  y  increase  and  decrease 
together. 

15.  Symbols  of  Abbreviation.  —  The  symbols  of  abbre- 
viation are  the  following : 

The  signs  of  deduction,  .*.  is  read  hence  or  therefore, 
and  •.*  is  read  since  or  because. 

The  signs  of  aggregation  are  the  bar  |  ,  the  parenthesis  (  ) , 
the  bracket  [  ] ,  the  brace  \  \ ,  and  the  vinculum  .  These 
are  employed  to  connect  two  or  more  numbers  which  are  to 
be  treated  as  if  they  formed  one  number.  Thus,  suppose 
we  have  to  denote  that  the  sum  of  a  and  b  is  to  be  multiplied 
by  c;  we  denote  it  thus  (a  +  b)  X  c  or  \a  4-  b\  x  c,  or 
simply  (a  +  b)  c  or  \a  4-  b\  c;  here  we  mean  that  the 
whole  of  a  +  6  is  to  be  multiplied  by  c.       If  we  omit  the 


ALGEBRAIC   EXPRESSIONS.  9 

parenthesis,  or  brace,  we  have  a  +  be,  and  this  denotes  that 
b  only  is  to  be  multiplied  by  c  and  the  result  added  to  a. 

Also  (a  +  b  +  c)  X  (d  -f  e)  denotes  that  the  result 
expressed  by  a  -f-  b  +  c  is  to  be  multiplied  by  the  result  ex- 
pressed by  d  -f-  e-  This  ma}'  also  be  denoted  simply  thus 
(a  -f-  b  -f  c)  (d  -f-  e),  just  as  a  X  b  is  shortened  into  ab. 
If  we  omit  the  parenthesis  we  have  a  +  b  +  cd  +  e,  and 
this  denotes  that  c  only  is  to  be  multiplied  by  d  o?J?/,  and  the 
result  added  to  a  -f-  6  -f  e. 

Also  V7  (a  -f  6  +  c)  denotes  that  we  are  to  obtain  the 
result  expressed  by  a  -f-  b  -f-  c,  and  then  take  the  square 
root  of  this  result. 

Also  (ab)'2  denotes  ab  x  ab  ;  and  (ab)s  denotes  ab  x  ab  x  a&. 

Also  (a  +  &  ■)-  c)  -r  (d  +  e)  denotes  that  the  result  ex- 
pressed by  a  +  b  +  c  is  to  be  divided  by  the  result  expressed 
by  d  -+-  e.  This  may  also  be  expressed  by  the  bracket  thus 
[a  -f-  b  +  c]  -5-  [d  +  e],  or  the  brace  ^a  +  6  +  c\  -r-  J  cZ  +  ef , 

a  +   6   +  c 

or  the  vinculum  a  -f-  6  -f-  c  -5-  d  -f-  e,  or ^7 -,  where 

the  line  between  the  numerator  and  denominator  acts  as  a 
vinculum. 

The  signs  of  continuation  are  dots ,  or  dashes 

,  and  are  read  and  so  on. 

16.  Algebraic  Expressions.  —  The  four  kinds  of  sym- 
bols which  have  been  explained  are  called  Algebraic  symbols 
(Art.  5).  An}'  collection  of  Algebraic  symbols  is  called 
an  Algebraic  expression,  or  briefly,  an  expression.  Thus 
4a  4-  5b  —  c  -f  x  is  an  expression  ;  3b  -f-  Ac  is  the  Alge- 
braic expression  for  3  times  the  number  b  increased  by  4 
times  the  number  c. 

The  numerical  value  of  an  expression  is  the  number 
obtained  by  giving  a  particular  value  to  each  letter,  and 
then  performing  the  operations  indicated. 

We  shall  now  give  some  examples  in  finding  the  numerical 
values  of  expressions,  as  an  exercise  in  the  use  of  the 
symbols  which  have  been  explained. 


10  EXAMPLES. 

EXAMPLES. 
If  a  =  1,  b  =  2,  c  =  3,  d  =  4,  e  =  5,  find  the  numerical 
values  of  the  following  expressions  : 

1.  9a  +  26  +  3c  -  2d. 

Here  we  have  9a  -f-  26  +  3c  —  2d  = 

9x1+2x2  +  3x3-2x4  = 
9_|_4  +  9_S  =  U     Ans. 

2.  7ac  +  36c  +  9d. 

Here  we  have  lae  +  36c  +  9d  = 

7x1x5  +  3x2x3 +  9x4  =  89  Ans. 

3.  «6cd  +  abce  +  «6de  +  acde  +  6cde.  Ans.  274. 

.     Aac    ,    86c       5cd  A 

4.  —  +  — .  t>. 

6  d  e 

f>     cde       56cd  _  Cade  g^ 
a6          ae            6c 

If  a  =  1,  6  =  3,  c  =  5,  and  d  =  0,  find  the  numerical 
values  of  the  following  : 

6.  a2  +  262  +  3c2  +  4d2.  Ans.  94. 

7.  a4  -  4a36  +  0a262  -  4«63  +  64.  16. 
12a3  -  62           2c2 a  +  62  +  c8                            5 

3a2  a  +  62  563 

If   a  =  1,   6  =  2,   c  =  3,   d  =  5,   and    e  =  8,    find    the 
numerical  values  of   the  following : 

9.    62(a2  +  e2  -  c2).  Ans.  224, 

10.  ^(26  +  -Id  +  5e).  8 

11.  (a2  +  Ir  +  c-)(e2  -  d*  -  c2).  420 

12.  e-  JV^(c  + l)+2;  +  (c-ye)V/(c-l).  15. 

13.  Find  the  value  of  x'1  —  2x  —  9  when  x  =  5. 

Explanation.  —  If  .r  =  any  number,  as  for  example,  ">,  then  xQ 

(which  =  x-  x)  =  5.r,  a-3  (which  =  x  -a*'-)  =  5j'2,  a*4  (which  =  .c  -x;!)  =  5x°, 
etc.    Hence  examples  like  i-'<  may  be  Bolved  as  follows: 

EXPLANATION. 

x1  —  2x  —   9  when  x  =  5.  x-  =  fee 

a;2  =  5x  n.r  —  2.r  =  ?>x 

Sx  = ,15  3.r  =  15 

0  =  resuh.  3x  —  9  =  15  —  9  =  ft. 


FACTOR  —  COEFFICIENT.  11 

14.  Find  the  value  of  or5-  50a;4-  49a;3-  lOOor-  101a?  -  50, 
when  x  =  51. 

These  examples  may  be  conveniently  solved  as  follows: 
xs  —  50.t4  —  49x3  —  100.1--  -  1013  -  50 
51  +  51      +  51      +  102     +  102     +  51 

+      z4  +    2x3  +      2x2  +       x  +    1         .-.     result  is  1. 

15.  Find  the  value  of  xA  —  liar3  —  liar8  —  11.?;  —  11  for 
x  =  12.  Ans.  1. 

1G.  Find  the  value  of  x*  -  8x3  -  19a;2  -  9a;  -  8  for 
X  =  10.  Ans.  2. 

17.  Factor  —  Coefficient.  —  When  two  or  more  num- 
bers are  multiplied  together  the  result  is  called  the  product, 
and  each  of  the  numbers  multiplied  together  to  form  the 
product  is  called  a  factor  of  the  product  (Art.  10).  Thus, 
3  x  4  x  5  =  GO,  and  each  of  the  numbers  3,  4,  and  5  is  a 
factor  of  the  product  60.  Factors  expressed  by  letters  are 
called  literal  factors ;  factors  expressed  by  figures  are  called 
numerical  factors.  Thus,  in  the  product  \ab,  4  is  called  a 
numerical  factor,  while  a  and  b  are  called  literal  factors. 

The  proof  is  given  in  Arithmetic  that  it  is  immaterial  in 
what  order  the  factors  of  a  product  are  written  ;  it  is  usual, 
however,  to  arrange  them  in  alphabetic  order. 

The  numerical  factor  is  called  the  coefficient  of  the  remain- 
ing factors.  Thus  in  the  expression  \ab,  4  is  the  coefficient, 
and  denotes  that  ab  is  taken  4  times.  But  it  is  sometimes 
convenient  to  consider  any  factor,  or  factors,  of  a  product 
as  the  coefficient  of  the  remaining  factors.  Thus,  in  the 
product  5a6c,  5a  may  be  appropriately  called  the  coefficient 
of  6c,  or  bab  the  coefficient  of  c. 

The  coefficient  is  called  numerical  or  literal,  according  as  it 
is  a  number,  or  one  or  more  letters.  Thus,  in  the  quantities 
5a;  and  mx,  5  is  a  numerical  and  m  a  literal  coefficient. 

When  no  numerical  coefficient  is  expressed,  1  is  always 
understood.     Thus,  a  is  the  same  as  la. 

A  coefficient  placed  before  any  parenthesis  indicates  that 


12  A    TERM,    ITS   DIMENSIONS,   AND    DEGREE. 

every  term  of  the  expression  within  the  parenthesis  is  to  he 
multiplied  by  that  coefficient. 

Care  must  be  taken  to  distinguish  between  a  coefficient  and 
an  exponent.  Thus  4a  means  four  times  a,  or  a  +  a  +  a  4-  ci ; 
here  4  is  a  coefficient.  But  a4  means  a  times  a  times  a  times  a, 
or  ax  ax  a  x  a,  or  aaaa  (Art.  12).     That  is,  if  a  =  4, 

4a  =  4  x  a  =  4  x  4  =  16, 

but    a4  =  a  x  a  x  a  X  a  =  4  x  4  x  4  x  4  =  256- 

18.  A  Term,  its  Dimensions,  and  Degree  —  Homo- 
geneous —  Similar.  —  A  term  is  an  Algebraic  expression 
in  which  no  two  of  the  parts  are  connected  by  the  sign  of 
addition  or  subtraction.  Thus  4a,  5a26c,  and  \xy  -r-  bob  are 
terms.  2a,  4c'2d,  and  —  bazd  are  the  terms  of  the  expression 
2a  +  4c2d  -  bahl. 

Each  of  the  literal  factors  of  a  term  is  called  a  dimension 
of  the  term,  and  the  number  of  the  literal  factors  or 
dimensions  is  called  the  degree  of  the  term.  Thus  a2b3c 
or  aabbbc  is  said  to  be  of  six  dimensions  or  of  the  sixth 
degree,  because  it  contains  six  literal  factors,  viz.,  a  twice, 
b  three  times,  and  c  once.  A  numerical  coefficient  is  not 
counted  ;  thus  a~b3  and  b<rbs  are  of  the  same  degree,  i.e.,  the 
fifth  degree,  since  there  are  five  literal  factors,  viz.,  a  twice 
and  b  three  times. 

It  is  clear  that  the  degree  of  a  term,  or  the  number  of  its 
dimensions,  is  the  sum  of  the  exponents  of  its  literal  factors, 
provided  we  remember  that  if  no  exponent  be  expressed  1 
must  be  understood  (Art.  12).  Thus  a*b*C?  is  of  the  ninth 
degree,  since  3  4-44-2  =  9. 

Terms  are  homogeneous  when  they  are  of  the  same  degree. 
Thus  a*6,  <t-b'\  Vuih'K  are  homogeneous. 

Terms  are  similar  or  like  when  they  have  the  same  literal 
part,  i.e.,  when  they  have  the  same  letters  and  the  corre- 
sponding letters  affected  with  the  same  exponents.  Other- 
wise they  are  said  to  be  unlike.      Thus,  the  terms  Sa'ft*, 

i)a'b\  and  —  a'^b'1  are  similar,  or  like  J   but  the  terms  <rb  and 


SIMPLE   AND    COMPOUND   EXPRESSIONS.  13 

ab2  are  unlike,  since,  although  the  letters  are  the  same,  they 
are  not  raised  to  the  same  power. 

19,  Simple  and  Compound  Expressions.  —  A  simple 
expression  consists  of  only  one  term,  as  oab,  and  is  called  a 
monomial. 

A  compound  expression  consists  of  two  or  more  terms,  and 
is  called  a  polynomial,  or  multinomial. 

A  binomial  is  a  polynomial  of  two  terms.  Thus,  ab2  +  2ac 
is  a  binomial. 

A  trinomial  is  a  polynomial  of  three  terms.  Thus,  a  -\-b  —  c 
is  a  trinomial. 

A  polynomial  is  said  to  be'  liomogeneous  when  all  its  terms 
are  of  the  same  degree.  Thus  oab2  +  ld2b  +  963  is  homo- 
geneous, for  each  term  is  of  the  third  degree. 

When  a  polynomial  consists  of  several  terms  of  different 
degrees,  the  degree  of  the  polynomial  is  that  of  its  highest 
term. 

A  polynomial  is  said  to  be  arranged  according  to  the 
powers  of  any  letter  it  contains  when  the  exponents  of  that 
letter  occur  in  the  order  of  their  magnitudes,  either  increasing 
or  decreasing.  Thus,  a4  +  4a36  +  Gcr62  +  4a&3  is  arranged 
according  to  the  descending  powers  of  a,  and  4a63  +  Ga2b2 
-f-  4a36  +  a4  is  arranged  according  to  the  ascending  powers 
of  a. 

The  reciprocal  of  a  number  is  1  divided  by  that  number. 

Thus,  the  reciprocal  of  a  is  -.     If  the  product  of  two  num- 

a 

bers  is  1,  each  number  is  the  reciprocal  of  the  other. 

20.  Positive  and  Negative  Quantities.  —  In  Arith- 
metic we  deal  with  numbers  connected  by  the  signs  -f-  and 
— ,  and  in  finding  the  value  of  an  expression  such  as 
14-3  —  2  +  4  we  understand  that  the  numbers  to  which 
the  sign  +  is  prefixed  are  additive,  and  those  to  which  the 
sign  —  is  prefixed  are  subtractive,  while  the  first  term,  1, 
to  which  no  sign  is  prefixed,  is  counted  among  the  additive 
terms.      The  same   thing  is  true  in  Algebra ;    thus  in   the 


14  POSITIVE  AND  NEGATIVE    QUANTITIES. 

expression  5a  +  76  —  3c  —  '2d  we  understand  the  sj'mbols 
5a  and  76  to  be  additive,  while  3c  and  2d  are  subtractive. 

But  in  Arithmetic  the  sum  of  the  additive  terms  is  always 
greater  than  the  sum  of  the  subtractive  terms,  i.e.,  we  are 
always  required  to  subtract  a  smaller  number  from  a  greater  ; 
if  the  reverse  were  the  case  the  result  would  have  no  Arith- 
metic meaning,  i.e.,  we  could  not  in  Arithmetic  subtract  a 
greater  number  from  a  smaller.  In  Algebra,  however,  not 
only  may  the  sum  of  the  subtractive  terms  exceed  that  of 
the  additive,  but  a  subtractive  term  ma}7  stand  alone,  and  yet 
have  a  meaniug  quite  intelligible.  It  is  therefore  usual  to 
divide  all  Algebraic  quantities  into  positive  quantities  and 
negative  quantities,  according  as  they  are  preceded  by  the 
sign  +  or  the  sign  —  ;  and  this  is  quite  irrespective  of  any 
actual  process  of  addition  and  subtraction. 

Illustration  (1)  Suppose  a  ship  were  to  start  from  the 
equator  and  sail  northward  100  miles  and  then  southward 
80  miles,  the  Algebraic  statement  would  be 

100  -  80  a  +20. 

Here  the  positive  sign  of  the  result  indicates  that  the  ship  is 
20  miles  north  of  the  equator.  But  if  the  ship  first  sailed 
80  miles  northward  and  then  southward  100  miles,  the  Alge- 
braic statement  would  be 

80  -  100  =   -20. 

clere  the  negative  sign  of  the  result  indicates  that  the  ship 
is  20  miles  south  of  the  equator. 

(2)  Suppose  a  man  were  to  gain  840  and  then  lose  $36, 
his  total  gain  would  be  $4.  But  if  he  first  gained  §30  and 
then  lost  $40,  he  sustained  a  loss  of  $1. 

The  corresponding  Algebraic  statements  would  be 

$40  -  $36  =  +81, 
$36  -  $40  =  -$4. 

Here  the  negative  quantity  in  the  second  case  is  interpreted 

as  a  debt,  i.e.,  a  sum  of  money  opposite  in  character  to  the 


POSITIVE   AND   NEGATIVE    QUANTITIES.  15 

positive  quantity  or  gain  in  the  first  case.  In  Arithmetic 
we  would  call  it  a  debt  or  loss  of  S4.  In  Algebra  we  make 
the  equivalent  statement  that  it  is  a  gain  of  —  84. 

(3)  Suppose  a  man  starts  at  a  certain  point  and  walks 
100  yards  to  the  right  in  a  straight  line,  and  then  walks  back 
70  yards,  he  will  be  30  yards  to  the  right  of  his  starting 
point.  If  he  first  walks  from  the  same  point  70  yards  to 
the  right  and  then  walks  back  70  yards,  he  will  be  at  the 
point  from  which  he  started.  But  if  he  first  walks  to  the 
right  70  yards  and  then  walks  back  100  yards,  he  will  be 
30  yards  to  the  left  of  his  starting  point.  The  corresponding 
Algebraic  statements  are 

100  yards  —     70  yards  =       30  yards 
70      "      —    70     "      =         0      " 
70      "      -  100      "      =  -30      "     . 
Here  we  see  that  the  negative  sign  may  be  taken  as  indi- 
cating a  reversal  of  direction.     In  Arithmetic  we  would  say 
the  man  was  30  yards  to  the  left  of  his  starting  point.     In 
Algebra  we  say  he  was  —30  yards  to  the  right  of  his  start- 
ing point. 

There  are  numerous  instances  like  the  preceding  in  which  it  is  con- 
venient for  us  to  be  able  to  represent  not  only  the  magnitude  but  the 
nature  or  quality  of  the  things  about  which  we  are  reasoning.  As  in 
the  preceding  cases,  in  a  question  of  position  we  may  have  to  distin- 
guish a  distance  measured  to  the  north  of  the  equator  from  a  distance 
measured  to  the  south  of  it;  or  a  distance  measured  to  the  right  of  a 
certain  starting  point  from  a  distance  measured  to  the  left  of  it;  or 
we  may  have  to  distinguish  a  sum  of  money  gained  from  a  sum  of 
money  lost ;  and  so  on.  These  pairs  of  related  quantities  the  A^e- 
braist  distinguishes  by  means  of  the  signs  -f  and  — .  Thus  if  the 
things  to  be  distinguished  are  gain  and  loss,  he  may  denote  by  4  or  +4 
a  gain,  and  then  he  will  denote  by  —4  a  loss  of  the  same  extent.  In 
this  way  we  can  conceive  the  possibility  of  the  independent  existence 
of  negative  quantities.  The  signs  +  and  — ,  therefore,  are  used  to 
indicate  the  nature  of  quantities  as  positive  or  negative,  as  well  as 
to  indicate  addition  and  subtraction  (Arts.  8  and  9). 

In  Arithmetic  we  are  concerned  only  with  the  numbers 
which  begin  at  0  and  are  represented  by  the  symbols  0,  1. 


16  POSITIVE  AND   NEGATIVE    QUANTITIES. 

2,  3,  4,  etc.  without  limit,  and  intermediate  fractions.  But 
the  quantities  which  we  usually  measure  by  numbers  in  Alge- 
bra do  not  really  begin  at  any  point,  but  extend  in  opposite 
directions  without  limit.  In  order  therefore  to  measure  such 
quantities  on  a  uniform  system,  the  symbols  of  Algebra  are 
considered  as  increasing  from  0  in  two  opposite  directions ; 
i.e.,  besides  the  symbols  used  in  Arithmetic,  we  consider 
another  set  —  1,  —2,  —3,  —4,  etc.  without  limit,  and  inter- 
mediate fractions.  Symbols  in  one  direction  are  preceded 
by  the  sign  +  ,  and  are  called  positive;  and  those  in  the 
other  direction  are  preceded  by  the  sign  — ,  and  are  called 
negative.  Symbols  without  a  sign  prefixed  are  considered 
to  have  -f-  prefixed. 

These  two  sets  of  symbols  may  be  illustrated  as  follows  : 

...  -8,  -7,  -6,  -5,  -4,  -3,  -2,  -1,     0.    +1,  +2,  +3,  +4,  +5,  +6,  +7.  +8,  .  .  . 
1 1 1 1 1 ! I i ! I I l l l I I I 

I 

the  positive  being  those  in  the  right  direction  from  zero, 
and  the  negative  those  in  the  left  direction  from  the  same 
point. 

Thus,  if  4  represent  a  distance  of  4  miles  measured  to  the  right  of 
a  certain  point,  —4  will  represent  a  distance  of  4  miles  measured  to 
the  left  of  the  same  point.  If  +4  represent  4  degrees  above  zero,  —4 
will  represent  4  degrees  below  zero.  If  +4  represent  4  years  after 
Christ,  —4  will  represent  4  years  before  Christ.  If  +4  represent  a 
fall  of  four  feet,  —4  will  represent  a  rise  of  4  feet.  If  +4  represent 
a  gain  of  $4,  —4  will  represent  a  loss  of  $4.  In  general,  when  we 
have  to  consider  quantities  the  exact  reverse  of  each  other  in  their 
nature  or  quality,  we  may  regard  the  quantities  of  either  quality  as 
positive,  and  those  of  the  opposite  quality  as  negative.  It  matters 
not  which  quality  we  take  as  the  positive  one  so  long  as  we  take  the 
opposite  one  as  negative;  but  having  assumed  at  the  commencement 
of  an  investigation  a  certain  quality  as  positive,  the  important  point 
is  to  use  it  uniformly  and  consistently  throughout. 

The  absolute  value  of  any  quantity  is  the  number  repre- 
sented by  this  quantity  taken  independently  of  the  sign 
which  precedes  the  number.  Thus,  3  and  —3  have  Uk 
-aine   absolute  value. 


MULTIPLICATIONS   IN  ANY    ORDER.  17 

Negative  quantities  are  often  spoken  of  as  less  than  zero. 
For  example,  if  a  man's  debts  exceed  his  assets  by  $4,  it 
is  said  that  "  he  is  worth  S-i  less  than  nothing."  In  the 
language  of  Algebra  it  would  be  said  "  he  is  worth  —  S-4." 

A  negative  number  is  said  to  be  Algebraically  greater  than 
another  when  it  is  numerically  less,  or  when  it  lias  the  smaller 
absolute  value.  Thus  —3  >  —  6,  since  —3  is  only  3  less 
than  0  while  —6  is  6  less  than  0,  or  as  a  person  who  owes 
$3  is  better  off  than  one  who  owes  $6  ;  or  in  the  case  of  the 
thermometer,  when  the  mercury  is  at  10°  below  0  (marked 
—  10°)  at  one  hour,  and  at  —  5°  at  another  hour,  the 
temperature  is  said  to  be  increasing;  i.e.,  —  5°  >  — 10°. 
Also,  in  Algebra,  zero  is  greater  than  any  negative  quantity, 
as  a  man  who  has  no  property  or  debt  is  considered  better 
off  than  one  who  is  in  debt.  Thus  it  is  easy  to  see  that 
in  the  series  on  page  16  each  number  is  greater  by  unity  than 
the  one  immediately  to  the  left  of  it. 

21.  Additions  and  Multiplications  may  be  Made 
in  any  Order. —  (1)  When  a  number  of  terms  are  con- 
nected by  the  signs  +  and  — ,  the  value  of  the  result  is  the 
same  in  whatever  order  the  terms  are  taken  ;  thus  6  +  5 
and  5  +  6  give  the  s'ame  result  viz.,  11  ;  and  so  also  a  +  b 
and  b  +  a  give  the  same  result,  viz.,  the  sum  of  the  num- 
bers which  are  represented  by  a  and  b.  We  may  express 
this  fact  Algebraically  thus,  a  +  b  =  b  +  a.  Similarly 
a  —  b  +  c  =  a  +  c  —  6,  for  in  the  first  of  the  two  expres- 
sions b  is  taken  from  o,  and  c  added  to  the  result;  in  the 
second  c  is  added  to  a,  and  b  taken  from  the  result. 

Similar  reasoning  applies  to  all  Algebraic  expressions. 
Hence  we  may  write  the  terms  of  an  expression  in  any  order 
we  please,  provided  each  has  its  proper  sign. 

Thus  it  appears  that  a  —  b  may  be  written  in  the  equiva- 
lent form  —  b  +  a.  As  an  illustration  we  may  suppose 
that  a  represents  a  gain  of  a  pounds,  and  —b  a  loss  of  b 
pounds  ;  it  is  clearly  immaterial  whether  the  gain  precedes 
the  loss,  or  the  loss  precedes  the  gain 


18  SUGGESTIONS  FOR   THE   STUDENT. 

(2)  When  one  number,  whether  integral  or  fractional,  is 
multiplied  by  a  second,  the  result  is  the  same  as  when  the 
second  is  multiplied  by  the  first. 

The  proof  for  whole  numbers  is  as  follows :  Write  down 
a  rows  of  units,  putting  b  units  in  each  row,  thus : 

I    |    |    |    | b  in  a  row, 

I    I    I    I    I 

I    |    |    |    | a  rows. 

Then  counting  by  rows  there  will  be  b  units  in  a  row  repeated 
a  times,  i.e.,  b  x  a  units.  Counting  by  columns  there  will 
be  a  units  in  a  column  repeated  b  times,  i.e.,  a  X  b  units. 

.-.  ba  =  ab. 

These  two  laws  are  together  called  the  Commutative  Laiv, 
or  Laic  of  Commutation. 

22.  Suggestions  for  the  Student  in  Solving  Ex- 
amples.—  In  solving  examples  the  student  should  clearly 
explain  how  each  step  follows  from  the  one  before  it ;  for 
this  purpose  short  verbal  explanations  are  often  necessaiy. 

The  sign  "  =  "  should  never  be  used  except  to  connect 
quantities  which  are  equal.  Beginners  should  be  particularly 
careful  not  to  employ  the  sign  of  equality  in  any  vague  and 
inexact  sense.  The  signs  of  equality,  in  the  several  steps 
of  the  work,  should  be  placed  one  under  the  other,  unless 
the  expressions  are  very  short. 

In  elementary  work  too  much  importance  cannot  be  at- 
tached to  neatness  of  style  and  arrangement.  The  beginner 
should  remember  that  neatness  is  in  itself  conducive  to 
accuracy. 

EXAMPLES. 

Find  the  numerical  value  of  the  following  expressions, 
when  a  =  1,  b  =  2,  c  =  3,  d  =  4,  and  e  =  5. 

1.  a*2  +  b2  +  c2  +  d1  +  e2.  Ana,  55. 

2.  abc-  4-  bed1  -  ,1m2.  94. 

3.  e4  +  <k-/r  +  V  -   lc;i/>  -   \eb\  81. 


EXAMPLES.  19 


,     &V    ,    de       32  j       ,q 

4.    —-  +  --  —  _.  Ans.  12 

4a        6-        6* 


o. 


a*  +  4a»5  _+_  Cjcrb2  +  4a&«  +  64 


a3  +  3a26  +  3a62  +  b* 

6.  (a  +  6)  (6  +  c)  -  (b  +c)  (c+d)  +  (c+d)  (d+e) .       43 

7.  (a-26  +  3c)2-(o-2c  +  3d)2+(c-2d+3e)2.       72 

8.  V^(4c2  +  Sd2  +  e).  11. 

9.  V^e2  +  cl2  +  c2  -  a2).  7. 
If  a  =  8,  6  =  6,  c  =  1,  x  =  9,  y  =  4,  find  the  value  of 

11.  Find  the  difference  between  a6x  and  a  +  b  -f-  ar,  when 
a  =  5,  6  =  7,  and  a;  =12.  ^l»s.  39G. 

12.  When  a  =  3,  find  the  difference  between  a2  and  2a, 
a9  and  3a,  a4  and  4a,  a6  and  5a,  a6  and  6a. 

Ans.  3,  18,  G9,  228,  and  711. 

13.  Find  the  value  of  3^c  +  2«V/(2a  +  6  -  *),  when 
a  =  G,  6  =  5,  c  =  4,  a;  =  1.  ^l//..s-.  54. 

14.  Find  the  value  of  (9  -  y)  (x  -f-  1)  +  (x  +  5)  {y  +  7) 
—  112,  when  x  =  3  and  y  =  5.  -4?is.  0. 

Find  the  value  of 

15.  «*  -  11  a3  -  11.x-2  -  13.x  +  11  for  x  =12.  -1. 

16.  .x4  -  x*  -  4x2  -  Sx  -  5  for  x  =  3.  4. 

17.  x5  -  3x2  -  8  for  x  =  4.  90s. 

18.  3x4  -  60.t-3  +  54a;2  +  GOx  +  58  for  x  =  19.         115. 
Express  the  following  in  Algebraic  symbols: 

19.  Seven  times  a,  plus  the  third  power  of  b.        7a  -f-  b:i. 

20.  Six  times  the  cube  of  a  multiplied  by  the  square  of  b. 
diminished  by  the  square  of  c  multiplied  by  the  fourth  power 
of  d.  Ans.  6a862  -  c2d\ 

21.  3  into  #  minus  m  times  y,  divided  by  m  minus  n. 

Ans.  (3x  —  my)  -f-  (m  —  n). 

22.  Four  times  the  fourth  power  of  a,  diminished  by  six- 
times  the  cube  of  a  into  the  cube  of  &,  and  increased  by 
four  times  the  fourth  power  of  b.       Ans.  4a4  —  Gazb3  -f  ib\ 

I 


20  ADDITION  —  ALGEBRAIC  SUM, 


CHAPTER     II. 

ADDITION. 

23.  Addition  —  Algebraic  Sum.  —  Addition  in  Alge- 
bra is  the  process  of  finding  the  Algebraic  sum  of  several 
quantities.  The  Algebraic  sum  of  several  quantities  is 
their  aggregate  value,  and  it  is  usual  to  find  the  simplest 
equivalent  expression  for  it. 

It  is  convenient  to  make  three  cases  in  Addition  ;  (1)  when 
the  terms  to  be  added  are  like  (Art.  18),  and  have  like  signs; 
(2)  when  they  are  like  but  have  unlike  signs;  and  (3)  when 
they  are  unlike. 

24.  Case  1.  To  Add  Terms  which  are  Like  and 
have  Like  Signs.  —  Let  it  be  required  to  add  8a;2?/,  ±x2y, 
and  7x2y. 

Here  8x2y  is  x2y  taken  8  times,  4x2y  is  x2y  taken  4  times, 
and  7x2y  is  x2y  taken  7  times  ;  therefore  x2y  is  taken  in  all 
8  +  4  -j-  7  =  19  times,  and  hence  the  sum  is  10x2y. 

The  truth  of  this  icill  be  evident  to  the  beginner  when  he 
remembers  that  the  three  quantities  8  lbs.,  4  lbs.,  and  7  lbs., 
added  together,  give  19  lbs. 

Similarly  12ab  +  Sab  +  5ab  +  ab  =  21ab. 

Let  it  be  required  to  add  —  oab,  —7(d),  and  — 0o6. 

Here  —  oab  is  ah  taken  —3  times,  —  lab  is  ab  taken  —7 
times,  and  —  9ab  is  ab  taken  —  9  times  ;  therefore  ab  is  taken 
in  ail  —19  times,  and  hence  the  sum  is  —  ldab. 

The  truth  of  this  will  be  evident  from  the  consideration  that, 
if  a  sum  of  money  be  diminished,  successively  by  $3,  $7,  and 
$9,  it  is  diminished  altogether  by  Slit. 

Therefore,  to  add  like  terms  which  have  the  same  sign,  add 
the  numerical  coefficients,  prefix  the  common  sign,  and  annex 
the  common  symbols. 


TO  ADD   LIKE    TERMS    WITH    US  LIKE   SIGNS.  21 

For  example,  6a  -f  3a  -f-  a  +  7a  =  17a,  and  —  2ab  —  lab 
—9db  =  —ISab. 

25.  Case  2.  To  Add  Terms  which  are  Like,  but 
have  Unlike  Signs.  —  Let  it  be  required  to  add  9a  and 
-4a. 

Here  —4a  destroys  4  of  the  9  time  ,  a,  and  gives  when 
added  to  it,  5a.  This  is  usually  expressed  by  saying  —  4a 
will  cancel  +4a  in  the  term  9a,  and  leave  -f  5a  for  the  aggre- 
gate or  sum  of  the  two  terms. 

For  if  9a  denote  $9  which  a  man  has  in  his  possession, 
and  —4a  denote  a  debt  of  84,  then  the  aggregate  value  of 
his  money  is  $o. 

In  like  manner  if  it  be  required  to  add  8a,  —9a,  —a,  3a, 
4a,  —11a,  a,  we  find  the  sum  of  the  positive  terms  to  be 
16a,  and  the  sum  of  the  negative  terms  to  be  —21a;  now 
+  16a  will  cancel  —16a  in  the  term  —21a,  which  leaves  —5a 
for  the  aggregate  or  sum  of  the  terms. 

Therefore,  to  add  like  terms  which  have  not  all  the  same 
sign,  add  cdl  the  positive  numerical  coefficients  into  one  sum, 
and  all  the  negative  numerical  coefficients  into  another;  take 
the  difference  of  these  two  sums,  prefix  the  sign  of  the  greater, 
and  annex  the  common  symbols. 

For  example  la  —  3a -f- 11  a+ a  —  5a  —  2a  —  19a  —  10a  =  9a, 
and  5a/j  —  Gab-\-2ab  —  lab  —  3ao-f-4aa  =  lla6  —  16a6  =  —hab. 

We  need  not,  however,  strictly  adhere  to  this  rule,  for 
since  terms  may  be  added  or  subtracted  in  any  order  (Art. 
21),  we  may  choose  the  order  we  find  most  convenient. 

Thus,  in  the  last  example,  we  may  say  bob  added  to  —  Gab 
gives  —  ab;  adding  —  ab  to  -{-2ab  gives  -{-ab;  adding  +ab 
to  —  lab  gives  —Gab;  adding  —Gab  to  —  Sab  gives  —  dab; 
adding  —  dab  to  +4ab  gives  —6ab,  for  the  sum,  which  is  the 
same  as  was  found  by  the  rule. 

26.  Case  3.  To  Add  Terms  which  are  not  all 
Like  Terms.  —  Let  it  be  required  to  add  4a  +  56  —  1c 
-f-  3d,  3a  —  b  -f-  2c  +  bd,  9a  —  26  —  c  —  d,  and  —a  -f-  3b 
+  4c  —  3d  -j-  e. 


22      TO  ADD  TERMS  WHICH  ARE  NOT  ALL  LIKE  TERMS. 

It  is  convenient  to  arrange  the  terms  in  columns,  so  that 
like  terms  shall  stand  in  the  same  column  ;  and  then  add 
each  column,  beginning  with  that  on  the  left,  as  follows : 

4a  +  56  —  7c  +  3d 
3a  —  6  +2c  +rod 
da  —2b  —  c  —  d 
-a  +36  +  4c  -3d  +e 


15a  -f-56  -2c  +4d  +e 

Here  the  terms  4a,  3a,  9a,  and  —a  are  all  like  terms ;  the 
sum  of  the  positive  terms  is  IGa ;  there  is  one  negative  term, 
viz.,  —a,  so  that  the  sum  of  the  terms  in  the  first  column  by 
Art.  25  is  -f-15a;  the  sign  -f-  may  be  omitted  by  Art.  9. 
Similarly  56  -  b  —  2b  +  36  =  56,  —  1c  +  2c  —  c  +  4c  =  -2c, 
and  so  on  ;  there  being  no  term  similar  to  e,  it  is  connected 
to  the  other  terms  by  its  proper  sign. 

Therefore,  to  add  terms  which  are  not  all  like  terms,  add 
together  the  terms  which  are  like  terms,  by  the  ride  in  Case  2, 
and  set  down  the  other  terms  each  preceded  by  its  proper  sign. 

In  the  two  following  examples  the  terms  are  arranged 
suitably  in  columns. 

xs  +2x-  -  3x  +1  a2  +  a6  +  62  -c 

4.x3  +  7ar  +     x  —9  3a2  -3ab  -762 

-28*  +  x2  -  9x  +  8  4a2  +5a&  +962 


9a;2  —     x  —  1  9a2  —  c 

In  the  first  example  we  have  in  the  fust  column  a;8  -f-  4ar 
—  2xs  —  ox*  =  5a;3  —  5&3  =  0  ;  this  is  usually  expressed  by 
saying  the  terms  which  involve  x3  cancel  each  other. 

Similarly,  in  the  second  example,  the  terms  which  involve 
<ih  cancel  each  other;  and  also  those  which  involve  62  cancel 
each  other. 

27.  Remarks  on  Addition.  —  We  have  seen  that  when 
two  or  more  like  terms  are  to  be  added  together  they  may 
be  collected  into  a  single  term  (Arts.  24    and   25).      If,  how- 


REMARKS    ON  ADDITION — EXAMPLES.  ^3 

ever,  the  terms  are  unlike  they  cannot  be  collected.  Thus 
we  write  the  sum  of  a  and  b  in  the  form  a  +  6,  and  the  sum 
of  a  and  —b  in  the  form  a  —  b. 

From  the  foregoing  examples  it  will  be  observed  that  in 
Algebra  the  word  sum  is  used  in  a  wider  sense  than  in 
Arithmetic.  Thus,  in  the  language  of  Arithmetic,  a  —  b 
signifies  that  b  is  to  be  subtracted  from  a,  and  has  no  other 
meaning  ;  but  in  Algebra  it  also  means  the  sum  of  the  two 
quantities  a  and  —b  without  any  regard  to  the  relative 
magnitudes  of  a  and  b. 

When  quantities  are  connected  by  the  signs  -f-  and  — ,  the 
resulting  expression  is  called  their  Algebraic  sum.  Thus 
the  Algebraic  sum  of  12a  —  29«  -f-  14a  is  —3a. 

In  Algebra,  wherever  the  word  sum  is  used  without  an 
adjective,  the  Algebraic  sum  is  understood. 


EXAMPLES. 

1. 

2. 

3. 

4. 

Aax 

Gab 

2bx2 

2a2b 

box 

—  lab 

-3bx2 

-  a2b 

—  3ax 

-Sab 

-Vbx2 

lla2b 

2ax 

5ab 

Ibx1 

-5a2b 

—  lax 

-9ab 

2bx2 

4a2b 

ax 

2ab 

-4bx2 

-9a2b 

2ax 

-Gab 

-hbx2 

2a2b 

5. 

lx2 

-3xy 

+     x 

3x2 

-  f 

+  3x  - 

-  y 

-2x2 

+4xy  +  hy2 

—     X 

-zy 

-  Ixy 

-  f 

+   9x  - 

-by 

4x2 

+±y2 

-   2x 

12;/r  —  Gxy  +  ly2  -f-lOfl!  —8y 

Add  together  the  following  expressions  : 

6.  a+26— 3c,  -3n  +  b  +  2c,  2a-3b+c.  Ans.O. 

7.  3a  +  2b  —  c,  —  a+36  +  2c,  2a  —  b+3c.         4a  +  46+4c. 


?4  EXAMPLES. 

S.  —  3x+2y+z,  x—3y  +  2z1  2x+y—3z.                 Ans.  0. 

9.  -x+2y  +  3z,  3x— y+2z,  2x+3y-z.           4a;-f4y-f-4g. 

10.  4«H-3&  +  5c,  —  2a+3&— 8c,  a  —  6+c.  3a  +  55  —  2c. 

11.  -15a-196— 18c,  14a  +  14&  +  8c,  a  +  5&  +  9c.  -c. 

12.  25a— 156+c,  13a— 10&+4c,  a-f-20&-c.  39a— 5&-|-4c. 

13.  -16a-10&  +  5c,  10a  +  5&+c,  6a+56— c.  5c. 
In  adding  together  several  expressions  containing   terms 

with  different  powers  of  the  same  letter,  it  will  be  found 
convenient  to  arrange  all  the  expressions  in  ascending  or 
descending  powers  of  that  letter  (Art.  19). 

14.  3£3  +  7-5.v2,  2a;2- 8 -9a;,  4a- 2^+ 3s2. 
Arranging  the   terms  in  the  descending  powers  of  x,  we 

have 

2x2  -9a;  -8 

-2a;3  +  3a;2  +4a; 

a;3  —5a;  —1  Ans. 

15.  3a&2-2&8+a8,    5a2&-a&2-3a8,    8a8-f568,    9a26-2a8 

-\-ab2.  Ans.  3&8+3a&2+14a26+4a8. 

It  will  be  observed  that  this  answer  is  arranged  according 
to  descending  powers  of  6,  and  ascending  powers  of  a. 

16.  2x2-2xy  +  3y2,  4y2+5xy-2x2,  x2-2xy-Gy2. 

Ans.  x2-\-xy+y2. 

17.  a8-a2+3a,  3a3+4a2+8a,  5a8-6a2-lla. 

Ans.  9a3 -3a2. 

18.  x»+3x2y+Sxy2,  -3x2y-Gxy2-x\  3x*y+4xy*. 

Ans.  3x*y+xy* 

19.  a;3-2«a;2+tt2a;+a3,  x*+3ax*i  2a8— aa;2— 2a£ 

Ans.  a2a;4-3a8. 

20.  2a&  -  3aa;2  +  2a2a;,    12ab  +  Wax2  -  6a2x,   -8ab  +  ax* 
—  f)a'2x.  Ans.  ()<d)~ 9a2a;4-7aa;2-f-aas*. 

21.  ar+?/4-f-28,  -4a;2-523,  8a;2-7?/4-f-1023,  cy-0,?3. 

^lres.  5a;2. 

22.  B*-4afy+6afy«— 4oy8+y4,   4a^-12a?y+12av/8-  ^i 
Gxy-12xy*+(Sy\  4xy*-  1;>\  y\  Ans.  x\ 


SUBTRACTION  -  ALU  KB  RAW   DIFFERENCE.  25 


CHAPTER    III. 

SUBTRACTION. 

28.  Subtraction  —  Algebraic  Difference.  —  Subtrac- 
tion in  Algebra  is  the  process  of  finding  the  difference 
between  two  Algebraic  quantities. 

The  Algebraic  Difference  of  two  quantities  is  the  number 
of  units  which  must  be  added  to  one  in  order  to  produce  the 
other.  Thus,  what  is  the  difference  between  2  and  6  means 
"  how  many  units  added  to  2  will  make  6  "  ?  The  Difference 
is  sometimes  called  the  Remainder. 

The  Subtrahend  is  the  quantity  to  be  subtracted;  or  it  is 
the  one  from  which  we  measure.  Thus,  2  is  the  subtrahend 
in  the  above  example. 

The  Minuend  is  the  quantity  from  which  the  subtrahend  is 
taken  ;  or  it  is  the  one  to  which  we  measure.  Thus,  6  is  the 
minuend  in  the  above  example. 

If  the  minuend  is  Algebraicallv  greater  than  the  subtra- 
hend, the  difference  is  positive  (Art.  20). 

If  the  minuend  is  Algebraically  less  than  the  subtrahend, 
the  difference  is  negative. 

In  Arithmetic  we  cannot  subtract  a  greater  number  from 
a  less  one,  because  subtraction  in  Arithmetic  means  taking  n 
less  number  from  a  greater.  But  in  Algebra  there  is  no  such 
restriction,  because  Algebraic  subtraction  means  finding  a 
difference. 

29.  Rule  for  Algebraic  Subtraction.  —  Let  distances 
to  the  right  of  the  zero  point  be  called  positive,  and  those 
to  the  left  of  the  same  point  be  called  negative  (Fig.  1,  Art. 
20).  Also  call  measuring  toward  the  right  from  any  point 
positive,  and  measuring  toward  the  left  from  any  point 
negative. 


26  RULE  FOR  ALGEBRAIC  SUBTRACTION, 

Then  the  difference  between  2  and  0  means  either  how 
many  units  must  we  measure,  and  in  what  direction,  in  order 
to  pass  from  2  to  6  or  to  pass  from  6  to  2.  In  the  first 
case  we  begin  at  2  and  measure  four  uuits  to  the  right  and 
say  2  from  6  is  -f  4.  In  the  second  case  we  begin  at  G 
and  measure  four  units  to  the  left  and  say  6  from  2  is  —4. 
That  is,  if  we  subtract  2  from  6  the  difference  is  4  ;  but  if 
we  subtract  6  from  2  the  difference  is  —4. 

Also  to  find  the  difference  between  —1  and  +1,  we  may 
begin  at  —1  and  measure  2  units  to  the  right  and  get  +2, 
or  we  may  begin  at  +1  and  measure  2  units  to  the  left  and 
get  —2  ;  i.e.,  if  we  subtract  —1  from  +1  the  difference  is  +2, 
but  if  we  subtract  +1  from  —1  the  difference  is  —2. 

Similarly  the  difference  between  —2  and  —7  is  —5  or 
+  5,  according  as  we  measure  from  —2  toward  the  left 
to  —7  or  from  —7  toward  the  right  to  —2;  i.e.,  if  we 
subtract  —2  from  —7  the  remainder  is  —5,  but  if  we  sub- 
tract —  7  from  —2  the  remainder  is  +5.  And  also,  the 
difference  between  —6  and  +7  is  +13  or  —13  according 
as  we  measure  from  —6  to  +7  or  from  +7  to  —  G  ;  i.e.,  if 
we  subtract  —6  from  +  7  the  difference  is  13,  but  if  we 
subtract  7  from  —6  the  difference  is  —13. 

Hence  we  see  that  the  remainder  in  each  case  is  found  by 
changing  the  Algebraic  sign  of  the  subtrahend,  and  then 
adding  it  Algebraically  to  the  minuend. 

Otherwise  thus.  Suppose  we  have  to  take  9  +  5  from  1 G  ; 
the  result  is  the  same  as  if  we  first  take  9  from  1G,  and  then 
take  5  from  the  remainder ;  that  is,  the  result  is  denoted  b}' 
16  -  9  -  5. 

Thus  16  -  (9  +  5)  =  16  -  9  -  5. 

Here  we  enclose  9  +  5  in  parenthesis  in  the  first  expres- 
sion, because  we  are  to  take  the  whole  of  9  +  5  from  16 
(Art.  15). 

Suppose  we  have  to  take  9  —  5  from  16.  If  we  take  9 
from  10,  we  obtain  10  —  9;  but  we  have  thus  taken  too 
much  from  1G,  for  we  had  to  take,  not  9,  but  9  diminished 


RULE   FOR   ALGEBRAIC   SUBTRACTION.  27 

by  5.  Hence  we  must  increase  the  result  by  5  ;  aud  thus  we 
obtain  16  -  (0  —  5)  =  1G  —  9  +  5. 

Similarly,  16  -  (6  +  4  -  1)  =  16  -  6  -  4  +  1. 

In  like  manner  suppose  we  have  to  subtract  b  —  c  from  a. 
If  we  subtract  b  from  a,  we  obtain  a  —  b ;  but  we  have  thus 
taken  too  much  from  a,  for  we  are  required  to  take,  not  6, 
but  b  diminished  by  c.  Hence  we  must  increase  the  result 
a  —  b  by  c ;  and  thus  we  obtain  a  —  (6  —  c)  =  a  —  b  +  c 
for  the  true  remainder. 

Similarly,  a  —  (b  +  c  —  d)  =  a  —  b  —  c  -f-  d. 

Suppose  we  have  to  subtract  b  —  c  -f-  c I  —  e  from  a.  This 
is  the  same  thing  as  subtracting  b  -f-  d  —  c  —  e  from  a  (Art. 
21).  If  we  subtract  b  +  d  from  a,  we  obtain  a  —  b  —  d ; 
but  we  have  thus  taken  too  much  from  c/,  for  we  were  to 
take,  not  b  +  d,  but  fr  +  d  diminished  by  c  and  e.  Hence 
we  must  increase  the  result  by  c  -f  e,  and  thus  obtain 

a—  (b—c+d  —  e)  =  a  —  b  —  d+c+e  =  a—  b+c  —  d+e. 

From  considering  each  of  these  examples,  it  is  evident 
that  subtracting  a  positive  number  is  the  same  tiling  as  adding 
an  equal  negative  number,  and  also  that  subtracting  a  negative 
number  is  the  same  thing  as  adding  an  equal  positive  number. 
Therefore,  Algebraic  subtraction  is  equivalent  to  the  Alge- 
braic addition  of  a  number  with  the  opposite  Algebraic  sign. 

Hence  for  subtraction  we  have  the  following 

Rule. 
Change  the  signs  of  cdl  the  terms  in  the  subtrahend,  and 
then  add  the  result  to  the  minuend. 


EXAMPLES. 

1 .  Let  it  be  required  to  subtract  ox—y+z  from  4x—Sy-\-2z. 

Changing  the  signs  of  all  the  terms  in  the  subtrahend,  it 
stands  as  follows:  —  Sx  +  y  —  z.  Then  collecting  as  in 
addition,  we  have 

4x  —  oy  +  2z  —  ?>x  +  y  —  %  =  x  —  2y  +  z. 


•28  EXAMPLES. 

2.  From  3a4  -f  bxs  —  Gx2  —  Ix  +  5 

take  2a;4  -  2a3  -f  5a;2  -  6x  -  7. 
Changing  the  signs  of  all  the  terms  in  the  subtrahend,  and 
proceeding  as  in  addition,  we  have 

3a;4  +  5ar*  -     6a;2  -  7x  +    5 
-2x*.+  2xs-     5a;2  +  6a;  -f-     7 


■■■      \  a;4  +  7a;3  -  11a;2  -    x  +  12 

Rem.  —  The  beginner  may  solve  a  few  examples  by  actually  changing 
the  signs  of  the  subtrahend  and  going  through  the  operation  as  fully  as 
we  have  done  in  these  two  examples;  but  he  may  gradually  accustom 
himself  to  perform  the  subtraction  without  actually  changing  the  signs, 
but  merely  changing  them  mentally,  as  in  the  following  example. 

3.  From  8ab  +  lac  -f  2c2  take  bob  —  4ac  +  3c2  —  d. 

Writing  the  subtrahend  under  the  minuend  so  that  similar 

terms  shall  fall  in  the  same  column,  for  convenience  (Art. 

26),  we  have  0  ,    .      -       .    „  2 

n  Sab  +    lac  -f-  2ca 

5ab  —    Aac  +  3c2  —  d 
Sab  +  llac  —    c2  +  d 

Changing  the  sign  of  5ab  from  -f  to  —  and  adding  it  to 
8a6,  we  have  Sab  ;  in  like  manner,  changing  the  sign  of  —  4ac 
from  —  to  +  and  adding  it  to  lac,  we  have  ll«c;  also 
changing  the  sign  of  -f-3c2  from  +  to  —  and  adding  it  to 
2c2,  we  have  — c2;  changing  the  sign  of  —  d  and  adding  it, 
we  have  +d. 

Every  example  in  subtraction  may  be  verified  by  adding 
the  remainder  to  the  subtrahend ;  the  sum  will  be  equal  U 
the  minuend. 

30.  Remarks  on  Addition  and  Subtraction.  —  In 
Arithmetic  addition  always  produces  increase  and  subtrac- 
tion decrease;  but  in  Algebra  addition  may  produce  decrease 
and  subtraction  may  produce  increase.  Thus  in  Algebra  we 
may  add  —Aa  to  8a  and  obtain  the  Algebraic  sum  4a,  which 
is  smaller  than  Sa ;  or  we  may  subtract  —3a  from  ha  and 
obtain  the  Algebraic  difference  8a,  which  is  larger  thau  ba. 


EXAMPLES.  29 


EXAMPLES. 


1.  From  5x2  +    xy  -  By2 
Subtract  2a?  -f  8xy  —  ly2 

Remainder  3a?  —  Ixy  +  4y2 

2.  From  x*  -  2x*              -     9x  + 
Subtract  2^4              -  3a?  +     lx  - 


Remainder         —  x4  —  2x*  +  °5x2  —  M>x  4-12 
From 

3.  lox  +  Mty  —  18z  subtract  2x  —  8y  +  z. 

Ans.  13a;  +  18?/  —  lft?. 

4.  x  —  y  —  z  subtract  —  10a;  —  14_y  4-  15z. 

Ans.  11a:  4-  13#  -  163. 

5.  25a  —  166  —  18c  take  4«  —  36  4-  15c. 

Ans.  21a  -  156  -  33c. 
G.    yz  —  za;  4-  37/  take  —  .ry  4-  72  ~  zx-  ^XV' 

7.  —2a3  —  x2  —  3x  4-  2  take  a?  —  a;  4-  1. 

Ans.  -3a?  -  x2  -  2a?  +  1. 

8.  4a?  -  3a;  +  2  take  -ojr  4-  6a;  -  7.     9a?  -  9a;  4-  9- 

9.  x9  4-  1  la?  4-  4  take  8a?  —  5x  —  3.     a?  4-  3a?  +  5as  +  7. 

10.  -8aV  4-  5a?  +  15  take  9aV  -  8a?  -  5. 

.4ns.  -17aV  4-  13^2  4-  20. 

11.  |a?  —  §a#  —  |y2  take  —fa?  4-  xy  —  y2. 

Ans.  2a?  —  f.r?/  —  A?/2. 

13.  Ja?  -  \x  4-  J  take  \x  -  1  4-  K"-     ~¥'  ~  P  +  J- 

14.  fa?  —  fax-  take  J  -  Ja?  —  |a».  fa?  4-  Jfla;  —  £. 

15.  fa?  -  §xtf  -  y2  take  Ja??/  -  f/  -  \xf. 

Ans.  fa?  —  ±x2y  —  £?/2. 

31.  The  Use  of  Parentheses.*— A  parenthesis  indicates 
that  the  terms  enclosed  within  it  are  to  be  considered  as  one 
quantity  (Art.  15).     On  account  of  the  extensive  use  which 


*  Afl  the  bracket,  brace,  bar,  and  vinculum  all  have  the  fame  significance  as  the 
parenthesis  (Art.  15),  the  rules  for  their  removal  or  introduction  are  the  same. 


12.    -frr  —  fa  —  1  take  —fa'2  -f-  a  —  \.     fa2  - 


30  MINUS   SIGN   BEFORE   THE   PARENTHESIS. 

is  made  of  parentheses  in  Algebra,  it  is  necessary  that  the 
student  should  become  acquainted  with  the  rules  for  their 
removal  or  introduction. 

32.  Plus  Sign  before  the  Parenthesis.  —  When  a 
parenthesis  is  preceded  by  the  sign  +,  the  parenthesis  can  be 
removed  without  making  any  change  in  the  exjiression  loithin 
the  parenthesis. 

This  rule  has  already  been  illustrated  in  Arts.  25  and  26  ; 
it  is  in  fact  the  rule  for  addition. 

7  +  (12  -f-  4)  means  that  12  and  4  are  to  be  added  and 
their  sum  added  to  7.  It  is  clear  that  12  and  4  may  be 
added  separately  or  together  without  altering  the  result. 

Thus         7  +  (12  +  4)  =  7  +  12  +  4  =  23. 

Also  a  +  (b  +  c)  means  that  b  and  c  are  to  be  added 
together  and  their  sum  added  to  a. 

Thus  a  +  (b  +  c)  =  a  -f  b  -f  c. 

7  +  (12  —  4)  means  that  to  7  we  are  to  add  the  excess 
of  12  over  4  ;  now  if  we  add  12  to  7,  we  have  added  4  too 
much,  and  must  therefore  take  4  from  the  result. 

Thus         7  +  (12  -  4)  =  7  +  12  -  4  =  15. 

Similarly  a  +  (b  —  c)  means  that  to  a  we  are  to  add  b 
diminished  by  c. 

Thus  a  -f-  (b  —  c)  =  a  -f-  b  —  c. 

Therefore  Conversely  :  Any  part  of  an  expression  may  be 
enclosed  within  a  parenthesis  and  the  sign  -f-  placed  before  it, 
the  sign  of  every  term  within  the  parenthesis  remaining  un- 
altered. 

Thus,  the  expression  a  —  b  +  c  —  d  +  e  may  be  written 
in  any  of  the  following  ways : 

a-b+c+(  —  d+e),  a— &+(c— d+e),  a+(— &-f  c— d+e), 

and  so  on. 

33.  Minus  Sign  before  the  Parenthesis.—  When  a 
parenthesis  is  preceded  by  the  sign  — ,  the  parenthesis  may 
be  removed  if  the  sign  of  every  term  within  the  parenthesis  be 
changed. 


COMPOUND    PARENTHESES.  31 

This  rule  has  already  been  illustrated  in  Art.  29  ;  it  is  in 
fact  the  rule  for  subtraction.  The  rule  is  evident,  because 
the  sign  —  before  a  parenthesis  shows  that  the  whole  ex- 
pression within  the  parenthesis  is  to  be  subtracted,  and  the 
subtraction  is  effected  by  changing  the  signs  of  all  the  terms 
of  the  expression  to  be  subtracted. 

Thus  a  —  (6  +  c)  =  a  —  b  —  c. 

Also  a  —  (b  —  c)  =  a  —  b  -f-  c. 

Therefore  Conversely:  Any  part  of  an  expression  may  be 
enclosed  within  a  parenthesis  and  the  sign  —  placed  before  it, 
provided  the  sign  of  every  term' within  the  parotthesis  be 
changed.  The  proof  of  this  operation  is  to  clear  the  paren- 
thesis introduced,  and  thus  obtain  the  original  expression. 

Thus  a  —  b  -\-  c  +  d  —  e  may  be  written  in  the  following 
ways : 

a— b-\-c—  (— cZ-f-e),  a— b—  (— c— cZ-f e),  a— (6— c— c/-(-e), 

and  so  on. 

34.  Compound  Parentheses.  —  Expressions  may  occur 
with  more  than  one  pair  of  parentheses ;  these  parentheses 
may  be  removed  in  succession  by  the  preceding  rules. 

We  may  either  begin  with  the  outside  parenthesis  and  go 
inward,  or  begin  with  the  inside  parenthesis  and  go  outward. 
It  is  usually  best  to  begin  with  the  inside  parenthesis.  The 
beginner  is  recommended  always  to  remove  first  the  inside 
pair,  next  the  inside  of  all  that  remain,  and  so  on.  Thus 
for  example  ; 

a  -f-  \b  +  (c  —  d)  \  =  a  +  \b  -f  c  -  d\  =  a  +  b  +  c  -  0 
a  -f  \b  —  (c  -  d)  \  =  a  -f  \b  —  c  +  d\  =  a  +  b  -  c  +  d. 
«  -  \b  +  (c  -  d)  \  =  a  -  \b  -f  c  -  d\  =  a  -  b  —  c  +  d. 
a  —  \b  —  (c  —  d)  I  =  a  —  \b  —  c  +  d\  =  a  —  b  -f-  c  —  d. 

It  will  be  seen  in  these  examples  that,  to  prevent  confusion 
between  different  pairs  of  parentheses,  we  employ  those  of 
different  forms;  and  hence  we  use,  besides  the  parenthesis, 
the  brace,  the  bracket,  and  sometimes  the  vinculum  (Art.  15). 


32  COMPO UND   PARENTHESES  —  EXAMPLES. 

Thus,  for  example, 

a  _  p  _  ic  _  (d  _  F=7)  j]  =  a  -  [6  -  f  c  -  (d  -  e  +  /)  ft 

=  a  -  [6  -  \c  -  d  +  e  -  J\]  =  a  -  [6  -  c  +  d  -  e  +  /] 
=  a  —  6  +  c  —  d  +  e—  /. 
Also 


i -2b-  [4a  -  66  -  §3a  -  c  +  (5a  -  26  -  3a  -  c  +  26)  j] 
=  tt-  26 -[4a -  66 - \3a  -  c  +  (5a  -  26  -  Za  +  c  -  26)  j] 

=  a  —  26  -  [4a  -  66  -  J3a  -  c  +  5a  -  26  -  3a  +  c  -  26  j] 
=  a  —  26  -  [4a  -  66  -  3a  +  c  -  ba  +  26  +  3a  -  c  +  26] 
=  a  _  26  —  4a  -f  66  +  3a  —  c  +  5a  —  26  —  3a  -+■  c  —  26 
=  2a,  by  collecting  like  terms. 

EXAMPLES. 

Simplify  the  following  expressions  by  removing  the  paren- 
theses and  collecting  like  terms. 

1.  a  — (6  — c)  +  a-f  (6  — c)+6— (a+c).    Ans.  a+6— c. 

2.  a  —  [6  +  \a  —  (6  +  a)  J],  a. 

3.  a  -  [2a  -  {36  -  (4c  -  2a)\].  a  +  36  -  4c. 

4.  $a_(6-c)$  +  f6-(c-a)}-{c-(a-6)|.    3a-6-c. 

5.  2a -(56  + [3c -a])  —  (5a—  [6+c]).     -2a— 46— 2c. 

6.  -[a-j6-(c-a)H-[6-Jc-(a-6)n.     6  -  a. 

7.  _(_(_(_a;)))_(_(_2/)).  a._  ?/. 

8.  -p.i;  -  (11t/  -  3a;)]  -  [by  -  (Sx  -  6y)].       -5*. 

9.  -[15a;  -  \Uy  -  (15s  +  12?/)  -  (10a;  -  15z)  J]. 

Ans.  — 25a;  +  2?/. 

10.  8a;-  jl6?/-[3^-(12?/-.T)-8?/]-f-.rj.     11a; -36?/. 

11.  -[a;  -  \z  +  (a;  -  z)  -  (a  -  a;)  -  z\  -  x],    2x  -  2z. 

12.  —[a  +  \a  -  (a  -  aj)  -  (a  +  a;)  —  a\  —  a].       2a. 

13.  — [a  —  ja  +  (a;  —  a)  —  (a;  —  a)  —  aj  —  2a].       a. 
1  i.    2a  -  [2a  -  \2a  -  (2a  -  2a  -  a)j].  a. 

15.  16  -  a;  -[7a;  -  \8x  -  (9x  -  3a;  —  6a;)|].  16  -  12a;. 

16.  2x  -  [8y  -  f4aj  -  (">//  -  6a?  -  7y)l].     12a?  -  15?/. 

17.  2a-[36  +  (2&-c)-4c  +  {2a-(36-c^26)|].   4c. 

18.  a-  [56-  \a-  (5c -2c- 6 -46)  +  2a  -  (a  -  26~+c)  j], 

^Lts.  oa  —  2c. 


EXAMTLES.  33 

19.  tT4_  r=4aj8_  jGa.2_  (4aj  _  1;  J]  _  (x*4  +  4a;3+r>.r2+4a;  +1). 

^4ns.  —  8x3  —  8x. 

When  the  beginner  has  had  a  little  practice  the  number 

of  steps  may  be  considerably  diminished ;    he  may  begin  at 

the  outside  and  remove  two  or  more  parentheses  at  once,  as 

follows : 

20.  a  -  [26  +  \3c  —  3a  —  (a  +  b) \  +  2a  -  (6  +  3c)] 
=    a  —  26  —  3c  +  3a  -f  a  +  6  —  2a  +  6  +  3c 

=  3a. 

21.  a-(6-c)  -  [a-6-c-2£6  +  c~-3(c-a)-dj]. 

2lns.  6a  -f  26  -  2c  -  2d. 

22.  2a; - (3y  -  4z)  -\2x  -  {Sy  +  \z)  \  -  \3y  - (4z  +  2a?)  \. 

Arts.  2x  —  3y  +  123. 

23.  -20  (a  -  d)  +  3(6  -  c)  -  2[6  +  c  +  d  -  3jc  +  d 

-  4(d  —  o)|].  Ans.  4a  -f-  6  +  c. 

24.  -4(a  -f  d)  +  24(6  -  c)  -  2[c  -f  d  +  a  -  3\d  +  a 

-  4(6  +  c)J].  Ans.  -50c. 

25.  2(36  -  5a)  -  7[a  -  GJ2  -  5(a  -  ft)  J]. 

^4?is.  -227a  -f  21G6  +  84. 

26.  -10Sa-6[a-(6-c)]J  +  G0j6-(c  +  a)J.     -10a. 

27.  _3j-2[-4(-a)]|  +  5|-2[-2(-a)]|.  4a, 

28.  -2\-[-(x  -  y)}\  +  \-2[-(x  -  y)]J.  0. 

Note.  —  The  line  between  the  numerator  and  denominator  of  a 
fraction  is  a  kind  of  vinculum.    Thus  — - —  is  equivalent  to  \(x  —  3). 

Ans.  -Va  —  26. 
30.    3^^^-^3x-5-(7x-iy)l']+8(y-  2a?). 

-4ns.  12a;  -  30y. 
3,|{|(o_6)_8(6-e)}-j^-^j 

-  ||c  -  a  -  |(a  -  6)  1.  -4ns.  a  -  ^6  +  Vc- 


34  EXAMPLES. 

-■  h!(f-!-)-HHH')HH-)} 

^4?is.  0. 
The  terms  of  an  expression  can  be  placed  in  parentheses 
in  various  ways  (Arts.  32  and  33).     Thus, 

33.  ax  —  bx  +  ex  —  ay  +  by  —  cy 
may  be  written 

(ax  -  bx)  +  (ess  —  ay)  +  (py  —  cy), 
or  (ax  —  bx  -f  ca;)  —  (a?/  —  6?/  +  c#), 

or  (ax  —  ay)  —  (6ic  —  by)  -f  (ca  —  c?/). 

Whenever  a  factor  is  common  to  every  term  within  a 
parenthesis,  it  may  be  placed  outside  of  the  parenthesis  as 
a  multiplier  of  the  expression  within.     Thus, 

34.  ax3  -+-  7  —  ex  —  dx2  —  c  -f-  6a;  —  dxz  -f-  6a;2  —  2x 
=  («a?  -  da;3)  +  (bx2  -  c7a;2)  +  (6a;  -  ex  -  2a?)  +  ( 7  -  c) 
=  (a  -  a")a?  +  (b  -  (Z)a;2  -f  (6  -  c  -  2)sc  +  (7  -  c). 

In  this  result,  (a  —  a*)*  (6  —  a*),  (6  —  c  —  2)  are  regarded 
as  the  coefficients  of  a;3,  a;2,  and  x,  respectively  (Art.  17). 
Hence  we  have  here  placed  together  in  parentheses  the 
coefficients  of  the  different  powers  of  x  so  as  to  have  the 
sign  +  before  each  parenthesis. 

35.  —  a2x  —  la  +  ahj  +  3  —  2a;  —  ab 

-  -(a2x  -  a2y)  -  (la  +  ab)  -  (2x  -  3) 
b-  -(a,  -  2/)a2  -  (7  +  6)a  -  (2a;  -  3). 
We  have  here  placed  together  in  parentheses  the  coeffi- 
cients of  the  different  powers  of  a  so  as  to  have  the  sign  — 
before  each  parenthesis. 

In  the  following  four  examples  place  together  in  paren- 
theses the  coefficients  of  the  different  powers  of  x  so  that 
the  sign  +  will  be  before  all  the  parentheses. 

36.  ax*  +  6a;2  +  5  +  26a;  -  5a£  +  2ar*  -  3a>. 

Ans.  (a  +  2)x*  +  (6  -  5)x*  +  (26  -  3)sc  +  5. 

37.  36a;2  -  7  -  2a;  +  cib  +  Sosc8  +  ca>  -  -l.r  -  6.t-3. 
Ans.  (5tt  -  6)a;3  +  (36  -  l)ar  +  (c  -  2)aJ  +  ab  -  7. 

38.  2  -  7a;3  +  5oa?  -  2ca;  +  9o#"  +  7a;  -  3a?. 

Ans.  (lJa  -  7).<;3  +  (5a  -  8)a*  -h  (7  -  2c)a; 


EXAMPLES.  35 

39.  2cx5  -  Sabx  4  idx  -  36a;4  -  aV  +  «*. 

J.ws.  (2c  -  a-)./;5  4(1-  36) x4  4  (4d  -  3a6)z. 
In  the  following  four  examples  place  together  in  paren- 
theses the  coefficients  of  the  different  powers  of  a;  so  that  the 
sign  —  will  be  before  all  the  parentheses. 
^40.    ax2  4  bx3  -  o¥  -  26a;3  -  3a;2  -  bx\ 

Ans.  -(a2  4  b)x*  -  (26  -  5)a;3  -  (3  -  a)x2. 

41.  7x3  —  oc^x  —  abx5  4  5cwj  4  ^"5  —  obex*. 

Ans.  —(ah  —  T)^5  —  («6c  —  l)x3  —  (3c2  —  5a)x. 

42.  aa;2  +  a^aj3  —  bx2  —  ox2  —  ex3. 

Ans.  —  (c  —  a2)^  —  (6  4  5  —  a)  a;2. 

43.  362a;4  -  fcc  -  ax4  -  ca-4  -  dc2x  -  7x\ 

Ans.  -(fl  +  c  +  7-  362)a;4  -  (6  +  5c2)a?. 

Simplify   the    following   expressions,   and  in  each  result 

place  together  in  parentheses  the  coefficients  of  the  different 

powers  of  x.     This  is  known  as  re-grouping  the  terms  accord- 

ing  to  the  powers  of  x. 

44.  as?-  2ex- [bx2-\cx-dx-  (bx3+3cx2)  \  -  (cx2-bx)]. 

Ans.  (a  -  b)x*  -  (6  4-  2c)x2  -  (b  +  c  4  d)x. 

45.  ax2  -  3\-ax3  4  36a;  -  4[£ca;3  -  %(ax  -  6a:2)]  j. 

Ans.  (3a  4-  2c):c3  4  (o  4  8o)a;2  -  (8a  4  96)x. 

40.  a;5-46.v4--(ri2ax-4 hbx*-d(™-  bx*\-^ax*\ 

Arts.  (66  4  l)*5  -  (a  426)a;4  -  (2a  4  3c) x. 
We  shall  close  this  chapter  with  a  few  examples  in  Addi- 
tion and  Subtraction. 

47.  To  the  sum  of  2a  -  36  -  2c  and  26  -  a  +  7c  add 
the  sum  of  a  —  4c  4  76  and  c  —  66.  Ans.  2a  4  2c. 

48.  Add  the  sum  of  2y  —  3y2  and  1  —  by3  to  the  remainder 
left  when  1  —  2y2  4  y  is  subtracted  from  by3.    Ans.  —y2  4  ?/• 

49.  Take  a;2  —  y2  from  3a;?/  —  4?/2,  and  add  the  remainder 
to  the  sum  of  4xy  —  x2  —  3y2  and  2a;2  -f-  6?/2.  ^4ns.  7a-?/. 

50.  Add  together  3x2  —  Ix  +  b  and  2x3  4  5a  —  3,  and 
diminish  the  result  by  3a;2  4  2.  Ans.  2X3  —  2x. 

51.  What  expression  must  be  subtracted  from  3a  —  56  4  c 
so  as  to  leave  2a  —  46  4  c?  -4ns.  a  —  6 


3G  MULTIPLICATION  —  RULE    OF  SJGNS. 


CHAPTER     IV. 

MULTIPLICATION. 

35.  Multiplication  in  Algebra  is  the  process  of  taking 
any  given  quantity  as  many  times  as  there  are  units  in  any 
given  number.* 

The  Multiplicand  is  the  quantity  to  be  taken  or  multiplied. 

The  Multiplier  is  the  number  by  which  it  is  multiplied. 

The  Product  is  the  result  of  the  operation. 

The  multiplicand  and  multiplier  taken  together  are  called 
Factors  of  the  product. 

In  Algebra  as  in  Arithmetic,  the  product  of  any  number  of 
factors  is  the  same  in  whatever  order  the  factors  may  be  taken 
(Art.  21).  Thus,  2x3x5  =  2x5x3  =  3x5x2, 
and  so  on.     In  like  manner  abc  =  acb  =  bca,  and  so  on. 

Also  2tt  x  36  =  2  x  a  X  3  x  b 

=2x3xaxl 
=  Gab. 

36.  Rule  of  Signs.  —  The  rule  of  signs,  and  especially 
the  use  of  the  negative  multiplier,  usually  presents  some 
difficulty  to  the  beginner. 

(1)  If  -fa  is  to  be  multiplied  by  -f-c,  this  indicates 
that  -fa  is  to  be  taken  as  man}'  times  as  there  are  units  in 
c.  Now  if  -fa  be  taken  once,  the  result  is  -\-a;  if  it  be 
taken  twice,  the  result  is  evidently  -f2a;  if  taken  three 
times,  the  result  is  -f  3a ;  and  so  on.  Therefore  if  -fa  be 
taken  c  times,  it  is  -\-ca  or  +ac.     That  is 

+a  x   +c  =  -f  ac. 

(2)  If  —a  is  to  be  multiplied  by  -f  c,  this  indicates  that 
—  a  is  to  be  taken   as  many  times  as  there  are  units  in  c. 


*  Thitt  definltioi)  i*  true  only  t>f  whole  muubcib. 


RULE    OF  SIGNS.  37 

Now   if    —a    be   taken    once,    the    result    is    —a;    if   it   be 
taken  twice,  the   result  is  —  '2a  ;    if   taken   three  times,  the 
result  is  —3a  ;    and  so  ou.      Therefore  if   —a  be    taken   c 
times,  it  is  —ca  or  —ac.     That  is, 
—  a  x   +c  =  —ac. 
►Similarly  —3  X    +  4  =  —3  taken  four  times 

=  -3  -3  -3  -3 
=  -12. 

(3)  Suppose  +a  is  to  be  multiplied  by  —  c.  "We  have 
illustrated  the  difference  between  +c  and  —  c  (Art.  20),  by 
supposing  that  +  c  represents  a  line  of  c  units  measured  in 
one  direction,  and  —  c  a  line  of  c  units  measured  in  the  oppo- 
site direction.  Hence  if  +«  is  to  be  multiplied  by  — e,  this 
indicates  that  -fa  is  to  be  taken  as  many  times  as  there  are 
units  in  +c,  and  further  that  the  direction  of  the  line  which 
represents  the  product  is  to  be  reversed. 

Now  +a  taken  +c  times  gives  -f-ac;  and  changing  the 
sign,  which  corresponds  to  a  reversal  of  direction,  we  get 
—ac.     That  is 

+  a  X   —  c  =  —ac. 

Similarly  +3  X  —  4  indicates  that  3  is  to  be  taken  4  times, 
and  the  sign  changed.  The  first  operation  gives  +12,  and 
the  second  —12.     That  is 

+3  x   -4  =  -12. 

(4)  If  —a  is  to  be  multiplied  by  — c,  this  indicates  that 
—  a  is  to  be  taken  as  many  times  as  there  are  units  in  c,  and 
then  that  the  direction  of  the  line  which  represents  the 
product  is  to  be  reversed. 

Now  —a  taken  c  times  gives  —  ac;  and  changing  the 
sign,  which  corresponds  to  a  reversal  of  direction,  we  get 
4-ac.     That  is 

—  a  x   —  c  =  +ac. 

Similarly  —3  x  —4  indicates  that  —3  is  to  be  taken  4 
times,  and  the  sign  changed.  The  first  operation  gives  —12, 
and  the  second  +12.      That  is, 

_3  X   -4  =   +12. 


38  EXAMPLES. 

(3)  is  sometimes  expressed  as  follows  :  +a  multiplied  by 
— c  indicates  that  -fa  is  to  be  taken  as  many  times  as  there 
are  units  in  e,  and  then  the  result  subtracted.  Now  -fa 
taken  +c  times  gives  -|-ac,  and  changing  the  sign,  in  order 
to  subtract  (Art.  29),  we  get  —  ac. 

Similarly  (4)  indicates  that  —a  is  to  be  taken  as  many 
times  as  there  are  units  in  c,  and  the  result  subtracted. 
Now  —  a  taken  +c  times  gives  —  ac,  and  changing  the  sign, 
in  order  to  subtract,  we  get  +ac. 

Hence  we  have  the  following  Rule  of  Signs :  The  product 
of  two  terms  with  like  signs  is  -f  ;  the  product  of  two  terms 
icith  unlike  signs  is  — . 

To  familiarize  the  beginner  with  the  rule  of  signs  we  add 
a  few  examples  in  substitution,  where  some  of  the  symbols 
denote  negative  quantities. 

EXAMPLES. 

If  a  =  -2,  b  =  3,  c  =  -1,  x  =  -5,  y  =  4,  find  the 
value  of  the  following  : 


1.   3a26  =  3(- 

-2)'2  x  3  = 

3 

X  4 

X  3  =  36. 

2.    —  7a8bc  =  - 

-7(-2)3  x 

3 

x  (- 

1)  =  -7  x  -8  X  3 

X   -1  =  -1G8. 

3.    Sabc2.         Ans.  -48. 

11. 

8cV.     Jxs.  -1000. 

4.    6aV. 

21. 

12. 

3c3.t-3.                    375. 

5.    —L2a4bx. 

480. 

13. 

hVJ.                    500. 

G.   5aV. 

500. 

14. 

7((\-\                 -224. 

7.    —7c*xy. 

140. 

ir>. 

_4oV.               -1G. 

8.    -Sax3. 

-2000. 

16. 

-6V.                   -9. 

9.    -5aW. 

-180. 

17. 

2a*c*x.                   40. 

10.    -7aV. 

-56. 

If  a  =  -4,  b 

n              

-hf 

=  0,  x  =  4,  ?/  =  1, 

find  the  value  of 

18.    3a'2  -f  bx  - 

-  icy. 

Ana.  40. 

19.    fa-  -  2&8  - 

-   c.r*. 

118. 

20.   Say  -  56s 

\c  -  2c3. 

-130. 

TIIE   MULTIPLICATION    OF  MONOMIALS.  39 

21.  2a3  -  Sbs  +  7nf.  Ans.  -54. 

22.  36V  -  4&!/"  -  Cc4x.  3. 

23.  2\T(ac)  -  B\f{xy)  +  VT^c4).  1. 

It  is  convenient  to  make  three  cases  in  Multiplication, 
(1)  the  multiplication  of  monomials,  (2)  the  multiplication 
of  a  polynomial  by  a  monomial,  and  (3)  the  multiplication  of 
polynomials. 

37.  The  Multiplication  of  Monomials.  —  Since  by 
definition  (Art.  12)  we  have 

a4  =  aaaa, 
and  a6  =  aaaaaa, 

.',     a4  x  aG  =  aaaaaaaaaa 
=  a10. 
Also  3a2  =  3aa, 

7«3  =  7aaa. 
.-.     3a2  x  7a3  =  3  x  7  x  aaaaa 
=  21a5. 
Similarly    5a362  x  6a*&V  =  haaabb  x  Gaabbbbxx 

=  30a5&6«2. 
•    Also    4a3c2  x  3c3£2  x  3#2  =  iaaaec  x  Scccxx  x  3rcaj 

==  30a3c5.r4. 
Hence  for  the  multiplication  of  monomials  we  have  the 
following 

Rule. 

Multiply  together  the  numerical  coefficients,  annex  to  the 
result  all  the  letters,  and  give  to  each  letter  an  exponent  equal 
to  the  sum  of  its  exponents  in  the  factors. 

For  example  2a;2  x  3a:4  x  x«  =  Gz2+4  +  6  =  Qx1'2. 

Also      5a2b3  x  262c4  x  3c2d4  =  30aWd4. 

Note.  —  The  beginner  must  be  careful,  in  applying  this  rule,  to 
observe  that  the  exponents  of  one  letter  cannot  combine  in  any  way 
with  those  of  another.  Thus,  the  expression  4a263c4  admits  of  no 
further  simplification. 

The  product  of  three  or  more  expressions  is  called  the 
continued  product. 


40  EXAMPLES. 

38.   To  Multiply  a  Polynomial  by  a  Monomial 

—  Suppose  we  have  to  multiply  (a  +  b)  by  3  ;  that  is,  take 
a  +  b  3  times.     We  have 

3(a  +  b)  =  (a  +  b)  4-  (a  +  &)  +  (a  4-  b) 
=  (a  +  (H   «  taken  three  times) 
together  with  (b  +  &  4-  b  takeu  three  times) 

=  3a  +  36. 
Similarly  7 (a  +  b)  =  la  +  lb. 

m(a  +  6)  =  (a  +  a  +  a  + taken  ??i  times) 

together  with  (b  +  b  -{- b  + takeu  m  times) 

=  ma  +  mb.  . (1) 

Also  m(a  —  b)  =  (a  +  a  -f-  a  + taken  m  times) 

together  with  (  —  b  —  b  —  b— taken  m  times) 

=  ma  —  mb (2) 

Similarly 

m(a  —  b  +  c)  =  ma  —  mb  -f  wic. 
This  is  generally  called  the  Distributive  Law. 
Hence,  to  multiply  a  polynomial  by  a  monomial,  we  have 

the  following 

Rule. 

Multiply  each  term  of  the  polynomial  separately  by  the* 
monomial,  and  collect  the  results  to  form  the  complete  product. 
For  example, 

4(x2  +  2xy  —  Az)  =  Ax2  +  8xy  —  lGz, 
(4^  _  7y  -  Sz3)  x  Sxy2  =  12x3y2  -  2lxy3  -  24a#¥*. 
(|aa  .     iab  _  jji)  x  6a262  =  4a4&3  -  a8&8  -  Ga2b\ 

EXAM  PLES. 

Multiply  together 

1.  4a268and7a8.  Am.  28a7&*. 

2.  3a4&V  and  5a8&«.  15a7W. 

3.  2.c2/y23  and  x'tfz.  2.v\>/8z*. 

4.  a6  +  be  and  a8&.  o46a  -h  «3b2c. 

5.  5aj  +  ."»//  and  2a8.  10.r3  +  6afy. 
be  +  Cf'  ~"  ^6  :ii id  n&c.  a//V2  -f-  o'dc8  —  <rb'2C. 

7.   5arfy  +  xy'1  —  7^ and  Sa?y.    *0aY  +  8»y  —  56»V 


THE    MULTIPLICATION   OF  POLYNOMIALS.  41 


8.    Ga3bc  —  7  aire2  and  a2b2. 

^Tis.  6a56*c  -  7a W, 

9.    a*b*x  -  bb%  and  2a8x5. 

2a76VJ  -  10aW, 

10.    Serb3  -  §62c3  and  |ct&*. 

12«367  -  «&6c3, 

39.  The  Multiplication  of  a  Polynomial  by  a 
Polynomial.  —  Suppose  we  have  to  multiply  c  -f  d  by 
a  4-  &. 

Here  we  are  required  to  take  c  +  d  as  many  times  as 
there  are  units  in  a  -f  &,  i.e.,  we  are  to  take  c  +  d  as  many 
times  as  there  are  units  in  a,  and  then  add  to  this  product 
c  -f  c I  taken  as  many  times  as  there  are  units  in  b. 

Hence  (a+b)  (c+d)  =  (c-\-d)  taken  a  times 
together  with  (c+d)  taken  b  times 

=  (c+d)a+(c+d)b 
=  «c+ad-f-6c+M[(l)ofArt.38].  (1) 
Again  (a  —  b)  (c-\-d)  =  (c+d)  taken  a  times 
diminished  by  (c+d)  taken  b  times 

=  (c-f-<Z)a—  (c-bd)ft 
=  ac+ad-  (bc+bd)  [(l)of  Art.  38] 
=  ac+ad— 6c— 6d (Art.  33).  .     .   (2) 
Also     (a+&)  (c—d)  =  (c—d)  taken  a  times 
together  with  (c—d)  taken  b  times 

=  (c— d)a-f-(c— d)6 
=  ac-ad+bc-bd[(2)of Art.38].  (3) 
Lastly  (a— 6)  (c—d)  =  (c— d)  taken  a  times 
diminished  by  (c  —  d)  taken  b  times 

=  (c—d)  a—  (c—d)b 
=  ac-ad-  (bc-bd)  [(2)  of  Art.  38] 
=  ac— ad — be + bd  (Art.  33 ) .  .     .   (4 ) 
Hence,  to  multiply  one  polynomial  by  another,  we  have 
the  following 

Rule. 

Multiply  each  term  of  the  multiplicand  by  each  term  of  the 
multiplier;  if  the  terms  multiplied  together  have  the  same  sign, 
prefix  the  sign  +  to  the  product,  if  unlike,  prefix  the  sign  —  ; 
then  add  the  partial  products  to  form  the  complete  product. 


42  EXAMPLES. 

If  we  consider  each  term  in  the  second  member  of  (4), 
and  the  way  it  was  produced,  we  lind  that 
+a  x  +c  =  +ac. 
-\-a  x  —  d  =  —  ad. 

—  b  X  +c  =  —  be. 

—  b  X  —  d  =  +bd. 

These  results  enable  us  again  to  state  the  rule  of  signs, 
and  furnish  us  with  another  proof  of  that  rule,  in  addition 
to  the  one  given  in  Art.  3G.  This  proof  of  the  rule  of  signs 
is  perhaps  a  little  more  satisfactory  than  the  one  given  in 
Art.  3G,  though  it  is  not  quite  so  simple. 

EXAMPLES. 

1.    Multiply  x  +  7  by  x  +  5. 
The  product  =  (x    +  7)  (x  +  5) 

=  x2  +  7x  -f-  5a5  +35 
=  x2  +  12a;  +  35. 

Rem.  —  It  is  more  convenient  to  write  the  multiplier  under  the 
multiplicand,  and  begin  on  the  left  and  work  to  the  right,  placing 
like  terms  of  the  partial  products  in  the  same  vertical  column,  ar 
follows : 

x  +    7 

x  +    5 


z2  +    7x 
+    5x  +  35 
by  addition  x2  +  12x  +  35. 


, 


2,   Multiply  3a;  -  4y  by  2x  -  3y. 
3a;    —     4y 
2x    —    By 

6a;2  —    8xy 

-     Oxy  +  V2if 


by  addition  Gar  —  Mxy  +  V2y*. 

Here  the  first  line  under  the  multiplier  is  the  product  of 
the  multiplicand  by  2a;;  the  second  line  is  the  product  of  the 


EXAMPLES.  43 

multiplicand  hf~—Sy\    like   terms  are   placed   in   the    same 
vertical  column  to  facilitate  addition. 
Find  the  product  of  the  following : 

Ans.  x2  —  17a;  4  70. 

x2  4  ?xx  -  70. 

x2  -  I3x  +  12. 

x2  -  225. 

x2  +  5a  4-  6. 

x2  -  25. 

a2  4-  x  -  306. 

a2  -  256. 

2x2  4-  13a;  -  24. 

6x2  4-  Us  -  35. 


3. 

a;  —  7  and  a;  —  10. 

4. 

x  —  1  and  .x*  4-  10. 

5, 

a;  —  12  and  x  —  1. 

6. 

x  —  15  and  a;  4-  15. 

7. 

—.v  —  2  and  —  x  —  3. 

8. 

— x  +  5  and  —x  —  5. 

0. 

a?  —  17  aud  x  4-  18. 

UK 

— x  —  1G  and  —a;  4  16. 

2a;  -  3  and  x  4-  8. 

§2> 

)  3a-  -  5  and  2a-  4-  7. 

l£ 

J  a-  -  5a&  4  662  and  2a2 

4a2  -    5a6    4-    662 

2a2  -    3a5    4-    4&2 

8a4  -  10«36  4-  12a262 

—  12«36  4-  15a262  -  18a&3 

4-  16a262  -  20abs  4  2164. 

8«4  -  22a36  4-  43a262  -  38a&3  4-  24&4. 


Here  the  first  line  under  the  multiplier  is  the  product  of 
the  multiplicand  by  2a2 ;  the  second  line  is  the  product 
by  —  Sab;  the  third  by  4&2 ;  like  terms  are  set  down  in  the 
same  vertical  column  to  facilitate  the  addition. 

The  student  will  observe  that  both  the  multiplicand  and 
multiplier  are  arranged  according  to  the  descending  powers 
of  a  (Art.  19).  Both  factors  might  have  been  arranged 
according  to  the  ascending  powers  of  a.  It  is  of  no  conse- 
quence which  order  we  adopt,  but  we  should  take  the  same 
order  for  the  multiplicand  and  multiplier. 

If  the  multiplier  and  multiplicand  are  not  arranged  accord- 
ing to  the  powers  of  some  common  letter,  it  will  be  convenient 
to  rearrange  them.     Thus  : 


-44  EXAMPLES. 

14.  Multiply  3*  +  4  -f-  2.r2  by  4  +  2x~  -  3a?.  Arranging 
the  factors  according  to  the  descending  powers  of  a;,  the 
operation  is  as  follows  : 

2s2  +  3x   +4 

2a;2  -  3x    +4 

4a;4  +  6a;3  +  8a2 

-  6a3  -  9a;2  -  12x 

+  8x*  +  12a  +  16 


4a;4  +  7a;2  +  16. 

15.   Multiply  a2  +  b2  4-  c2  -  aft  -  6c  —  cabya+fc  +  c 

Arrange  according  to  descending  powers  of  a. 
a'2  —  ah   —  ac    -\-  b2    —    be    +  c2 
a    4-  6     +  c 


a8 

— 

a26 

— 

a2c 

+ 

«62  - 

abc 

+ 

ac2 

+ 

a26 

— 

ab2  - 

abc 

4-  63 

-  b2c 

4-  &c2 

+ 

a2c 

— 

abc 

— 

ac2 

4-  b2c 

-  6c2 

+  c8 

a« 

— 

3abc 

4-  b3 

4-  c3 

Rem.  —  The  student  should  notice  that  he  can  make  two  exercises 
in  multiplication  from  every  example  in  which  the  multiplicand  and 
multiplier  are  different  polynomials,  by  changing  the  original  multi- 
plier into  the  multiplicand,  and  the  original  multiplicand  into  the 
multiplier.  The  result  obtained  should  be  the  same  in  both  opera- 
tions. The  student  can  therefore  test  the-  correctness  of  his  work  by 
interchanging  the  multiplicand  and  multiplier. 

Multiply  together 

16.  a2  -  ab  +  b2  and  a2  +  ab  4-  b2.     Ana.  a4  +  a2b2  +-  bA. 

17.  a;2  +  3?/2  and  x  +  Ay.  a;3  +  Axhj  +  3a;?/2  +  12//\ 

18.  x4  -  x2f  -f-  y4  and  a;2  4-  2/2-  a8  4-  ?A 

19.  a2  —  2ax  4-  4ar  and  a2  +  2ax  -f  4a;2.    a4  4-  4a2.c2  +  1 6aJ*. 

20.  IGa2  4-  12a6  4-  9&2  and  4a  -  3b.  04a3  -  276s. 

21.  a2x  —  ax2  4-  a;3  —  a3  and  a;  -f-  a.  a;4  —  a4. 

22.  2a;3  -  3x2  4-  2a;  and  2a;2  4-  3a;  4-  2.     4a;5  -  a;3  +  Ax. 

23.  -a5  4-  a*b  —  a86a  and  -a  —  o.  «fi  4-  a*b*. 

24.  a3  4-  2tr&  4-  tab'2  and  aa  -  2ab  4-  26*.       a5  4-  Aab\ 


EXAMPLES  45 

Whea  the  coefficients  are  fractional  we  use  the  ordinary 
process  of  multiplication,  combining  the  fractional  coefficients 
by  the  rules  of  Arithmetic* 

25.   Multiply  \a2  -  $ab  4  ^r  by  Ja  +  \b. 

\ab   +  |  b2 


**' 


& 


-Ja3  -    larb  +  \ab2 

+    herb  -  lab2  +  |68 


Ja3 

+ 

J5a 

-A- 

l^3  ' 

-If. 

lr4 
4  J/ 

— 

4  3r2 

+  *■ 

lax 

K 

+  a4. 

Ja3  -  3Vr6  +  Ja&2  +  | 

Multiply  together 

26.  -Ja2  +  Ja  +  J  and  |a  —  J.      ^4??s. 

27.  fa;2  +  xy  +  §?/2  and  \x  -  \y. 

28.  \x2  -  fa?  -  |  and  ^2  4  fa;  -  |. 

29.  |aaj  +  §;«2  +  -Ja2  and  fa2  +  far8  - 

it  is  sometimes  desirable  to  indicate  the  product  of  poly- 
nomials, by  enclosing  each  of  the  factors  in  a  parenthesis, 
and  writing  them  in  succession.  When  the  indicated  multi- 
plication has  been  actually  performed,  the  expression  is  said 
to  be  expanded,  or  developed. 

Expand  the  following : 

30.  (2a  +  36)  (a  -  b).  Ans.  2a2  4-  db  -  362. 

31.  (a2  +  ax  4-  x2)(a2  -  ax  4-  x2).         x*  +  a2x2  4  a\ 

32.  (a2  4  2db  +  2b2)  (a2  -  2ab  4  2b').  a4  +  4b\ 

33.  (x  -  3)(a  4-  4)0  -  5)0  4  6). 

Ans.  x4  +  2x3  -  41a2  -  42z  4  360 

34.  (a2  4  ab  4  62)  (a3  -  a26  4-  b3)  (a  -  6). 

-4ns.  a6  ~  a56  -f  a264  -  &6. 

35.  (2a3  +  4x2  +  8oj  4-  16)(3oj  -  6).  Go;4  -  9G. 
30,  0s  +  a;2  4-  x  -  \){x  -  1).  a4  -  2.x-  4  1. 
37.    O  4-  «)[0  +  b)(x  4  c)-  (a  +  6  +  c)(x  4  &) 

4-  {a2  +  a&  +  6'2)].  Ans.  x3  -f  a3. 

*  The  student  is  supposed  to  be  familiar  with  Arithmetic  fractions,  which  are  the 
only  fractions  that  are  used  in  this  work  previous  tc  Chapter  VIII. 


4G       MULTIPLICATION   BY   INSPECTION  —  EXAMPLES. 

40.  Multiplication  by  Inspection.  —  Although  the 
result  of  multiplying  together  two  binomial  factors  can 
always  be  obtained  by  the  methods  explained  in  Art.  39, 
yet  it  is  very  important  that  the  student  should  learn  to 
write  down  the  product  rapidly  by  inspection.  This  is  done 
by  observing  in  what  way  the  coellicients  of  the  terms  in  the 
product  arise ;  thus 

(a  -f  5)  (a  +  3)  =  a2  +  5a  +  3a  -f-  15 

==  x2  +  8a  -f-  15. 
(x  -  5)  (x  +  3)  =  a2  -  5a  +  3a  -  15 
=  x2  —  2x  —  15. 


(x  +  5)  (a 

_  3)  =  x2  +  5a  _  3a  -  15 

=  x2  +  2x  -  15. 

(a  —  5)  (x 

-  3)  =  x2  -  bx  -  3a  +  15 

=  ^  -  8a  +  15. 

It  will  be  noticed  in  each  of  these  results  that : 

1.  The  product  consists  of  three  terms. 

2.  The  first  term  is  a2,  and  the  last  term  is  the  product  of 
ihe  second  terms  of  the  two  binomial  factors. 

3.  The  middle  term  lias  for  its  coefficient  the  Algebraic 
turn  of  the  second  terms  of  the  two  binomial  factors. 

Hence  the  intermediate  steps  in  the  work  may  be  omitted, 

and  the  product  written  down  at  once,  as  follows : 

(a  +  2)  (a  +  3)  =  x2  +  5a  +    6. 

(a  -  3)  (a  +  4)  =  x2  +    a  -  12. 

(a  +  6)  (a  -  9)  =  x2  -  3a  -  54. 

(a>  _  4y)(x  -  10//)  =  a-  -  May  +  -10/. 

(a  —  by)  (a  +     Gy)  =  x2  +       xy  —  30/. 

EXAMPLES. 

Write  down  the  values  of  the  following  products. 
I      (a  +  8)  (a  -  5).  -4ns.  a3  +  3a  -  40. 

2.    va  -  3)  (a  +  10).  a'-  +  7a  -  30, 

ti.    (x  +  7)  (a  —  Si],  a"2  -  2a  -  63. 

4.    (a  —  4)  (a  +  II).  x1  +  7a  —  44. 


SPECIAL-FORMS    OF    MULTIPLICATION.  47 

5.  [x  +  2)(x-  5). 

6.  {x  -j-  9)(as  -  5). 

7.  (a;  -  8)(aj  +  4). 

8.  (as  -  6) (x  +  13). 

9.  (a  -  11) (as  +  12). 
10.  (x  -  3a)  (a;  +  2a). 

€£  (a5  -  96)(aj  +  86). 

3)  (*-  730(3 -8y). 


-4ns 

.  x2  -  3a; 

-  10. 

0^  +  43; 

-  45. 

x2  —  4a; 

-  32. 

a;2  +  7a; 

-  78. 

a;2  +  x  - 

-  132. 

x2  —  ax  - 

-  6a2. 

x2  —  6a;  — 

7262. 

X2 

—  Ihxy  +  b(jy2. 

4L  Special  Forms   of  Multiplication  —  Formulae. 

—  There  are  some  examples  in  multiplication  which  occur 
so  often  in  Algebraic  operations  that  they  deserve  especial 
notice. 

If  we  multiply  a  -f  6  by  a  -f-  6  we  get 

(a  +  6)  (a  +  6)    =  a2  +  2a6  +  62 ; 

that  is  (a  +  6)2  =  a2  +  2«6  -f-  62.  .     .     .    (1) 

Thus  the  square  of  the  sum  of  hvo  numbers  is  equal  to  the 
siim  of  the  squares  of  the  two  numbers  increased  by  twice  their 
product. 

Similarly,  if  we  multiply  a  —  6  by  a  —  6  we  get 

(a  -  b)2  =  a'2  -  2a6  +  62 (2) 

Thus  the  square  of  the  difference  of  two  numbers  is  equal 
to  the  sum  of  the  squares  of  the  two  numbers  diminished  by 
twice  their  product. 

Also,  if  we  multiply  a  -f  6  by  a  —  6  we  get 

(a  +  6)  (a  -  6)  =  a2  -  62 (3 J 

Thus  the  product  of  the  sum  and  difference  of  tico  numbers 
in  equal  to  the  difference  of  their  squares. 

Because  the  product  of  two  negative  factors  is  positive 
(Aiv.  3G),  it  follows  that  the  square  of  a  negative  number 
is  positive.     For  example, 

(-«)«=««  =  (+a)', 

and  (6  —  a)2  =  a2  —  2a6  +  62  =  (a  —  6)2. 


48  EXAMPLES. 

Hence  a2  —  2a b  +  b2  is  the  square  of  both  a  —  b  and 
&  —  a. 

Rem.  1.  —  Equations  (1),  (2),  and  (3)  furnish  simple  examples  of 
one  of  the  uses  of  Algebra,  which  is  to  prove  general  theorems 
respecting  numbers,  and  also  to  express  those  theorems  briefly. 

For  example,  the  result  (a  +  b)  (a  —  b)  =  a2  —  b2  is 
proved  to  be  true,  and  is  expressed  thus  by  symbols  mure 
compactly  than  it  could  be  by  words. 

A  general  result  thus  expressed  by  symbols  is  called  a 
formula;  hence  a  formula  is  an  Algebraic  expression  of  a 
general  rule. 

Rem.  2.  — We  may  here  indicate  the  meaning  of  the  sign  ±  which 
is  made  by  combining  the  signs  +  and  — ,  and  which  is  called  the 
double  sign. 

By  using  the  double  sign  we  ma}-  express  (1)  and  (2)  in 
one  formula  thus : 

(a  ±  b)2  =  a2  ±  2ab  +  b\   .     .     .     .   (4) 

where  ±,  read  plus  or  minus,  indicates  that  we  may  take 
the  sign  -f-  or  — ,  keeping  throughout  the  upper  sign  or  the 
lower  sign.  Formulae  (1),  (2),  and  (3)  are  true  whatever 
may  be  the  values  of  a  and  b. 

The  following  examples  will  illustrate  the  use  that  can  be 
made  of  formulae  (1),  (2),  and  (3).  The  formulae  will 
sometimes  be  of  use  in  Arithmetic  calculations.     Thus 


EXAMPLES. 

1.  Required  the  difference  of  the  squares  of  127  and  123. 
By  formula  (3)  we  have 

(127)2  -  (123)2  =  (127  +  123)  (127  -  123) 
=     250  x  4  =  1000. 

2.  Required  the  square  of  2(J. 
By  formula  (2) 

(29)a=  (30  -  1)-  =  <J00  -  60  +  1  =  Sil. 


EXAMPLES.  49 

3.  Required  the  product  of  53  by  47. 
By  formula  (3) 

53  x  47  =  (50  +  3)  (50  -  3)  =  (50)2  -  (3)3 
=    2500  -  9  =  2491. 

4.  Required  the  square  of  34. 
By  formula  (1) 

(34)2  =  (30  +  4)2  =  900  +  240  +  16  =  1156. 

5.  Required  the  square  of  4a  4-  3y. 

We  can  of  course  obtain  the  square  by  multiplying  4a  -f-  3y 
by  itself  in  the  ordinary  way.  But  we  can  obtain  it  by 
formula  (1)  more  easily,  by  putting  4a;  for  a  and  3y  fori. 
Thus 

(4a  +  Sy)2  =  (4a)2  +  2(4a%)  +  (3#)2 
=  16a2    +  24a>/        +  9y\ 

6.  Required  the  square  of  a  -f  y  +  z. 

Denote  a  +  y  by  a ;  then  x-\-y  +  z  =  a-\-z\  and  by 
(1)  we  have 

(a  +  z)2  =  a2  +  2az  +  z2 

=  (x  +  y)2  +  2(a?  -f  y)a  4-  z2 
=  a2  +  2a?/  +  y2  +  2az  -f  2^/z  4-  sg 
Thus  (a  4-  V  4-  z)2  =  x2  +  y2     +  z2  +  2xy  4-  2yz  +  2xz. 

That  is,  the  square  of  the  sum  of  three  numbers  is  equal 
to  the  sum  of  the  squares  of  the  three  numbers  increased  by 
tioice  the  products  of  the  three  numbers  taken  tivo  and  two. 

7.  Required  the  square  of  p  —  q  4-  r  —  s. 

Denote  p—q  by  a  and  r—s  by  b  ;  then  p  — q+r— s=a-\-b  ; 
and  by  (1)  we  have 

(a4-&)2=a24-2a&4-&2=  (p-Q)2+^(p-Q)  (»"-*)  +  (r-s)2. 
Then  by  (2)  we  expand  {p  —  q)2  and  (r  —  s)a. 

Thus  (p  —  q  4-  r  —  s)2 

=  }t  —  2pq  4-  (f-  +  2  ( ;?/•  —  ps  —  qr  4-  r/.s)  4-  ?-2—  2rs  4-  a2 
=  ^»'2  4-  q2  +  r2  H-  s2  4"  2^/'  4-  2(/.v  —  2pg  -  -  2ps  —  2qr  —  2rs. 


50  EXAMPLES. 

8.  Required  the  product  of  p— q-\-r— s  and  p— q— r-f-s. 
Let  p  —  q  =  a  and  r—s  -  b  ;  tben  p  —q  -j-r  —  s  =  a-f&, 

and  p  —  q  —  r  +  s  =  a  —  b;  and  by  (3)  we  have 

(a+6)  (o-6)  =a2-b2=  (p-q)2-  (r-s)2 

=  p2-2pq+q2~{r2-2rs+s2)  by  (2). 

Thus(p— g+r— s)(p— q— r+s)=p2-\-q2— r2— s2— 2pq-\-2rs. 

From  these  examples  we  see  that  by  using  formulae  (1), 
(2),  and  (3),  the  process  of  multiplication  may  be  often 
simplified.  The  student  is  advised  first  to  go  through  the 
work  fully  as  we  have  done ;  but  when  he  becomes  more 
familiar  with  this  subject,  he  may  dispense  with  some  of  the 
work,  and  thereby  simplify  the  multiplication  still  more. 
Thus  in  the  last  example  he  need  not  substitute  a  and  6,  but 
apply  formula  (3)  at  once,  and  then  (2),  as  follows  : 

[(P  -  9)  +  (r  -  *)]  [<J>  ~q)~  (r  -  •)]  =  (p  -  q)2 

—  (r  —  s)2  =  p2  —  2pq  +  q2  —  r2  -f  2rs  —  s2. 

9.  Required   the    product    of    a  4-  b  +  c,   a  4-  b  —  c, 
a  —  6  +  c,  6-fc  —  a. 

By  (3)  and  (1)  we  obtain  for  the  product  of  the  first  two 
factors, 

(a  +  b  +  c)  (a  +  b  —  c)  =  2ab  -f  a2  +  62  -  c2.  .   ^1) 

By  (3)  and  (2)  we  obtain  for  the  product  of  the  last  two 
factors, 

(a-b  +  c)  (b  +  c-a)  =  2ab  -  (a2  +  b2  -  c2).   .   (2) 
Multiplying  together  (1)  and  (2),  we  obtain 
(2a6)2  -  (a2  +  b2  -  c2)2 

=  2a262  +  262c2  +  2a2c2  -  a4  -  b4  -  c4.  .    .     .   (3) 

Solve  the  following  examples  in  multiplication  by  formulae 
(1),  (2),  and  (3).    * 

f0^)  (15x  +  14?/)2.  Ans.  225a;2  -f-  420.r?/  +  H-6?/2. 

Cm  (7a;2  -  by2)2.  49a;4  -  7().r//2  +  25y*. 

12.  (x2  +  2x  -  2)2.  a;4  +  lx»  -  8a;  --  4. 

13.  (x2  -  [)x  -f  7)2.  «4  -  IOjc8  +  39a?1  -  70a;  +  49. 


IMPORTANT  RESULTS   IN   MULTIPLICATION  51 

14.  (2x2  -  Sx  -4)2.  Ans.  4.x-4  -  12a*3  -  7x2  +  24aj  4  IB. 

15.  ( x  4  2y  +  3z) 2.  Xs  +  4y2  4  92s  +  4ajy  4  6a»  -f- 1 2?/z. 
1G.  (x-2  +  .n/  +  y2)  (a;2  4  a#  -  r) .  ^  +  2afy  4  x2y2  -y\ 
1 7.  (x2  4  ^  4  ?/2)  (x2  -  a^  4  r) .                    z4  4-  ^V  4-  t- 

42.  Important  Results  in  Multiplication.  —  There 
are  other  results  in  multiplication  which  are  important, 
although  they  are  not  so  much  so  as  the  three  formulae  in 
Art.  41.  We  place  them  here  in  order  that  the  student  may 
be  able  to  refer  to  them  when  they  are  wanted ;  they  can  be 
easily  verified  by  actual  multiplication. 

(a  4  6)  (a2  -  db  4  b2)  =  a3  +  b*.  .     .     .   (1) 

(a  -  b)  (a2  4  db  4  b2)  =  a3  -  b\  .     .     .  (2) 

(a 46)3  =  (a+b)  (a2  +  2a6  +  62)  =  a3  4-  3a26  +  3a&2  4  &3.   (3) 

(«-&)«= (a-&)  (a2-  2a&  4-  &2)  =  a3 -  3a26  +  3a&2 -  b\  (4) 

(a  4  b  4-  c)3  =  a8  4  3a*(6  4  c)  4  3a(6  4-  c)2  +  (b  +  c)8 

=  a34&34c343a2(64c)  +  36a(a+c)  +3c2(a4&)  +6o6c.   (5) 

Rem.  —  It  is  a  useful  exercise  in  multiplication  for  the  student  to 
show  that  two  expressions  agree  in  giving  the  same  result.  For 
example,  show  that 

(a  —  6)(6  -  c){c  -  a)  =  a2{c  -  b)  +  63(a  -  c)  4  c2(b  -  a). 
Here  we  proceed  as  follows:  Multiplying  (a  —  b)  by  (6  —  c)  we 
obtain 

(a  _  b)[b  -  c)  =  ab  -  b2  -  ac  +  be; 

then  multiplying  this  equation  by  c  —  a  we  obtain 

{a  -  b){b  -c){c-  a)  =  cab  —  cb2  -  ac2  4  be2  -  a2b  4  ab2  4  a2c  -  ahr 
=  a2(c  -  b)  4  b2(a  -  c)  4  c2(b  -  a)  .    .     .   (0; 

Show  that  (a  -  6)2  4  (6  -  c)2  4  (c  -  a)2 

=  2(c  -  6)(c  -  a)  +  2(6  -  o)(6  -  c)  4  2(a  -  b)(a  -  c), 
By  (2)  of  Art.  41  we  obtain 
(a  -  6)2  +  (6  -  c)2  +  (c  -  a)2 

=  a2  -  2a6  4  62  4  62  -  26c  +  c2  +  c2  -  2ac  +  a2 
=  2(a2  +  62  +  c2  —  a6  —  6c  —  ca) (7 ; 

Now  (c  —  6)(c  —  a)  =  c2  —  ca  —  cb  +  ab, 

(6  —  a)  (6  —  c)  =  6'2  —  6c  —  ab  +  ac, 

(a  —  6)  (a  —  c)  =  a2  —  «c  —  a6  +  6c; 


52    RESULTS    OF  MULTIPLYING    ALGEBRAIC  EXPRESSIONS. 

therefore,  by  adding  these  three  equations,  we  obtain 

(c  _  b){c  -  a)  +  (6  —  a)(b  -  c)  +  {a  —  b)(a  —  c) 

—  a2  +  b'2  +  c2  —  ab  —  ac  —  6c,    (8) 
therefore,  from  (7)  and  (8)  we  have 

(a  _  b)2  +  [b  -  c)2  +  (c  -  «)2 

=  2(c  —  fc)(c  —  a)  +  2{b  -  a)(b  -  c)  +  2(a  —  b)(a  —  c).     (9) 

43.  Results  of  Multiplying  Algebraic  Expressions. 

—  From  an  examination  of  the  examples  in  multiplication, 
the  student  will  recognize  the  truth  of  the  following  laws 
with  respect  to  the  result  of  multiplying  Algebraic  expres- 
sions. 

(1)  In  the  multiplication  of  two  polynomials ,  when  the 
'partial  products  do  not  contain  like  terms,  the  whole  number 
of  terms  in  the  final  product  will  be  equal  to  the  product 
of  the  number  of  terms  in  the  multiplicand  by  the  number  of 
terms  in  the  multiplier,  but  will  be  less  if  the  partial  products 
contain  like  terms,  owing  to  the  simplification  produced  by 
collecting  these  like  terms. 

Thus  as  we  see  in  Ex.  17,  Art.  39,  there  are  tivo  terms  in 
the  multiplicand  and  two  in  the  multiplier,  and  four  in  the 
product,  while  in  Ex.  13  there  are  three  terms  in  the  multi- 
plicand and  three  in  the  multiplier,  and  only  five  in  the 
product. 

(2)  Among  the  terms  of  the  product  there  are  ahvays  two 
that  are  unlike  any  other  terms;  these  are,  that  term  which  is 
the  product  of  the  two  terms  in  the  factors  which  contain 
the  highest  power  of  the  same  letter,  and  that  termiuhich  is  the 
product  of  the  two  terms  in  the  factors  which  contain  the  low- 
est power  of  the  same  letter. 

Thus  in  Ex.  13,  Art.  39,  there  are  the  terms  8a4  and  2464, 
and  these  are  unlike  any  other  terms ;  in  fact,  the  other 
terms  contain  a  raised  to  some  power  less  than  the  fourth, 
and  thus  they  differ  from  8a4 ;  and  they  also  contain  a  to 
some  power,  and  thus  they  differ  from  2464. 

(3)  Wlten  the  multiplicand  and  multiplier  are  both  homo- 
geneous (Art.  18)  the  product  is  homogeneous,  and  the  degree 


EXAMPLES.  53 

of  the  product  is  the  sum  of  the  numbers  which  exjjress  the 
degrees  of  the  multiplicand  and  multiplier. 

Thus  in  Ex.  13,  Art.  39,  the  multiplicand  and  multiplier 
are  each  homogeneous  and  of  the  second  degree,  and  the 
product  is  homogeneous  and  of  the  fourth  degree.  In  Ex. 
15,  Art.  39,  the  multiplicand  is  homogeneous  and  of  the 
second  degree,  and  the  multiplier  is  homogeneous  and  of  the 
first  degree ;  the  product  is  homogeneous  and  of  the  third 
degree.  This  law  is  of  great  importance,  as  it  serves  to  test 
the  accuracy  of  Algebraic  work  ;  the  student  is  therefore 
recommended  to  pa}7  great  attention  to  the  degree  of  the 
terms  in  the  results  which  he  obtains. 

EXAMPLES. 

Multiply 

1.  4a2  -  3b  by  Sab.  Ans.  12a3b  -  dab2. 

2.  8a2  -  dab  by  3a2.  24a4  -  27asb. 

3.  3x2  -  Ay2  +  bz2  by  2x2y.        6a;4?/  -  8x2y3  +  10x2yz2. 

4.  x2y3  -  y3z4  +  zAx2  by  x2y2z2.        x4y5z2  -  x2y5zG  -f  xYz6. 

5.  2xy2z3  +  3x2y3z  —  hx3yz2  by  2xy2z. 

Ans.  Axhfz4  -f-  Gx3y5z2  -  10x4y3z3. 

6.  -2a26  -  4«52by  -7a262.  14a463  -f  28a364. 

7.  8xyz  -  10x3yz3  by  -xyz.  -8x2y2z2  +  I0x4y2z4. 

8.  abc  —  a2bc  —  ab2c  by  —abc.     —a2b2c2  +  a362c2  +  a263c2. 

9.  x  -f  7  by  x  —  10.  x2  —  3x  —  70. 

10.  x  +  9  by  x  —  7.  x2  +  2x  —  63. 

11.  2x  -  3  by  x  -f  8.  2x2  +  13a;  -  24. 

12.  2x  +  3  by  x  -  8.  2a;2  -  13x  -  24. 

13.  x3  -  7x  -f-  5  by  x2  -  2x  +  3. 

Ans.  x5  -  2x4  -  4a;3  +  19a;2  -  31a  +  15. 

14.  a2  -  dab  -  b2  by  a2  +  5ab  +  b2. 

Ans.  a4  -  25a262  -  10a&3  -  64. 

15.  a;2  —  xy  +  x  +  y2  +  y  +  1  by  ac  +  ?/  -  1. 

Ans.  x3  -\-  3xy  +  y3  —  1. 

16.  a2  +  62  +  c2  --  6c  -  ca  -  a&  by  a  +  &  +  c. 

Ans.  a3  -j-  63  +  c3  —  3a6c 


54  EXAMPLES. 

1 7.  2ax+x2+a2  by  a2+2ax-x2.  Am.  a4-f-4a3a;-f  4a2a;2-a;4. 

18.  262+3a&-a2by7a-56.         -1063-a62  + 2  Ga26- 7a3. 

19.  a2-a6H-62bya2+a6-62.  a4-a2&2+2a&8-&4. 

20.  4a*-3a^-#2by3a>-2#.  12z3-17.v2?/+3a;?/2+2?/3. 

21.  a^-afy-fa^-^byaH-y.  x6-xy+xY-y\ 

22.  cc4+2^  +  4^24-8^3+lG^/4bya;-2?/.  a;5-32?/5. 

23.  9a;V2+27a;^  +  81^  +  3.T?/3+?/4by3x-2/.         243a;5-;?/5. 

24.  aj+2?f-3sbya;-2y-t-3z.  a;2-4?/2+12?/z-9z2. 
Write   down   the   values    of    the    following   products   by 

inspection. 

25.  (x  +  7)(aj  +  1).  -4n*.  ^,2  +  8^  +  7. 

26.  (a>  -  7)  (a;  +  14).  x2  +  7x  -  98. 

27.  (a  +  36)  (a  -  26).  a2  +  a&  -  6&2. 

28.  (a  -  G)(a  +  13).  a2  +  7a  -  78. 

29.  (2a;  -  5)(x  -  2).  2a;2  -  9a;  +  10. 

30.  (3a;  -  l)(a;  +  1).  3a;2  +  2a;  -  1. 

31.  (3a;  +  7) (2a;  -  3).  6a;2  +  6x  -  21. 
Solve  the  following  examples  by  formulae  (1),  (2),  (3)  in 

Art.  41. 

32.  (x2-\-xy-\-y2)(x2— xy— y2).      Ans.  x4—x2y2—2xys—y* 

33.  (a?+xy-y2)  (x2-xy+y2).  x*-x*y*+2xy*-tfi 

34.  (a;3+2a;2+3a'-f l)(.«3-2.u2+3a;-l).     o;6+2.r4+5a;2-l 

35.  (a;-3)2(a?+6aH-9).  a;4-18a;2+81 
3G.  (a;+2/)2(a-2-2.r?/-?/2).  a;4-4a'V2-4a^3-?/4 
Show  that  the  following  results  are  true : 

37.  (aa+62)  (c2+d2)  =  (ac  +  bd)2+  (ad-bc)2. 

38.  (a+b+c)*+a*+b*+c*=(a+by+(b+c)*+(c+a)\ 

39.  (a-b)  (b-c)  (c-a)=bc(c—b)  +ca(a-c)  +ab(b— a) 

40.  (a-6)8+68-a8=3a6(6-a). 

41.  (d2+ab  +  b2)2-(a2-ab  +  lr)2=\ab(a2+b2). 

42.  (a  +  b+c)s-as-bs-cs=X{a  +  b)(b  +  c)(c+a). 

43.  (a+6)2+2(a2-62)  +  (a-&)2=4aa. 

44.  (a— &)8+(&-c)8+(c-a)8=3(a-6)(6-c)(c-a). 


rlUE   DIVISION    OF   ONE    MONOMIAL   BY  ANOTHER.       55 


CHAPTER   V. 

DIVISION. 

44.  Division  in  Algebra  is  the  process  of  finding,  from 
a  given  product  and  one  of  its  factors,  the  other  factor ;  or 
it  is  the  process  of  finding  how  many  times  one  quantity  is 
contained  in  another. 

Division  is  therefore  the  converse  of  multiplication. 

The  Dividend  is  the  given  product ;  or  it  is  the  quantity 
to  be  divided. 

The  Divisor  is  the  given  factor ;  or  it  is  the  quantity  by 
which  we  divide. 

The  Quotient  is  the  required  factor ;  or  it  is  the  number 
which  shows  how  many  times  the  divisor  is  contained  in  the 
dividend. 

The  above  definitions  may  be  briefly  written 

quotient  x  divisor  =  dividend, 

or  dividend  -r-  divisor  =  quotient. 

It  is  sometimes  better  to  express  this  last  result  as  a 

fraction ;  thus  dividend  , . 

=  quotient. 

divisor 

It  is  convenient  to  make  three  cases  in  Division,  (1)  the 
division  of  one  monomial  by  another,  (2)  the  division  of  a 
polynomial  by  a  monomial,  (3)  the  division  of  one  poly- 
nomial by  another. 

45.  The  Division  of  one  Monomial  by  Another. 
—  Since  the  product  of  4  and  x  is  4a;,  it  follows  that  when 
4#  is  to  be  divided  by  x  the  quotient  is  4.     Or  otherwise 

4iB  -5-  X  =  4. 

Also  since  the  product  of  a  and  b  is  ab,  the  quotient  of  ab 
divided  by  a  is  b  ;  that  is 

ab  -j-  a  =  6. 


")G      THE   DIVISION    OF    ONE    MONOMIAL    EY  ANOTHER. 

Similarly  abc  -r-  a  =  be ;    abc  -f-  5  ==  ac  ;    abc  -f-  c  =  ab  \ 

aba  -r-  ab  =  c  ;  abc  -f-  be  =  a  ;  a&c  -i-  ca  =  6.     These  results 

may  also  be  written 

abc  _  ,    m     abc  _       ,     abc  _    ,  , 

abc 

abc  abc  abc       , 

— -  =    c;     —  =    a;     —  =&. 
ao  oc  ca 

. ,       0„  „        n  4         36a6       SGaaaaaa       .        , 

Also  36a6  -^  9a4  =    = =  4aa,  by  remov- 

9a4  daaaa 

ing  from  the  divisor  and  dividend  the  factors  common  to 

both,  just  as  in  Arithmetic. 

Therefore  36a6  --  9a4  =  4a2. 

<-,..,,  AK  4i o  o        n  n.5       45aaaabbbcc 

Similarly        4oa4crcJ  -f-  9a~6c-  = 

daabec 

Hence  we  have  the  following 

Rule. 

To  divide  one  monomial  by  another,  divide  the  coefficient  of 
the  dividend  by  that  of  the  divisor,  and  subtract  the  exponent 
of  any  letter  in  the  divisor  from  the  exponent  of  that  letter  in 
the  dividend. 

For  example    72xby3  -v-  Ux*y2  =  Ga?~Y~* 

=  Gx2y. 

Also  55a4x3?/5  -r-  lla2xy2  =  &a*x*y*. 

Rem. — If  the  numerical  coefficient,  or  the  literal  part  of  the 
divisor  be  not  found  in  the  dividend,  we  can  only  indicate  the  division. 
Thus  if  7a  is  to  be  divided  by  2c,  the  quotient  can  only  be  indicated 
by  7a  -f  2e  or  by—.     In  some  cases,  however,  we  may  simplify  the 

expression  for  the  quotient  by  a  principle  already  used  in  Arithmetic. 
Thus  if  ldcflb?  is  to  be  divided  by#12«6c,  the  quotient  is  denoted  by 

— - — -.     Here  the  dividend  =  4ab  x  4a26  and  the  divisor  =  -lab  x  3c; 
I2abc 

thus  the  factor  4ab,  which  occurs  in  both  dividend  and  divisor,  may 
be  removed  in  the  same  way  as  in  Arithmetic,  and  the  quotient  will  be 

denoted  by  ^L2.     That  is 
J    '3c 

lfia3h2  _  4ab  x  4a -b  _  4nn-b 
\2<lbc  An!)   x  8c  So 

by  removing  the  common  factor  4<ib. 


THE  RULE   OF   SIGNS  —  EXAMPLES.  .r>7 

Note.      If  we  apply  the  above  rule  to  divide  any  power  of  a  letter 
by  the  same  power  of  the  letter,  we  are  led  to  a  curious  conclusion. 
Thus  by  the  rule 

a3  -f  a3  =  a3-3  =  a0; 

but  also  «3  -r  a3  =  -}      =  1, 

a3 

by  removing  the  common  factor  a3; 

.".  o°  =  1; 
that  is,  any  quantity  whose  exponent  is  0  is  equal  to  1. 

The  true  significance  of  this  result  will  be  explained  in 
Art.  115. 

46.  The  Rule  of  Signs  for  division  may  be  obtained 
from  an  examination  of  the  cases  which  occur  in  multiplica- 
tion, since  the  product  of  the  divisor  and  quotient  must  be 
equal  to  the  dividend. 


s  we  have 

+a  x   (+6)  =  +ab,     . 

,-.   +ab  -s- 

+  5  =  -f-a. 

— a  X    (+6)  =  —  ab,     . 

\    -ah  -r- 

+  b  =   —a. 

-fa  x   (—6)  =  —  ah,     . 

,-.    -ah  ~- 

—b  =  +a. 

—  a  x   (— b)  =  -Ha&,     . 

•.    +«6  -T- 

—  h=  —a. 

Hence  in  division   as  well   as  in  multiplication,  like  signs 
produce  -f ,  and  unlike  signs  produce  —. 

EXAMPLES. 

Divide 

1.  8a2b  by  4Mb.  Ans.  2a. 

2.  —\hxy  by  3a;.  — 5#. 

3.  -21a263by  -7a2h\  36. 

4.  45a<Wby  -9a8&ar*.  —haW. 
24aW.  fa&2. 

3.r. 

29a262c2. 

-9.i'5. 

2zm. 


5. 

-36aWby  -i 

C. 

3xs  by  x2. 

7. 

-58aWby  -2 

8. 

63aV#2  by  —7a' 

9. 

4zw+n  by  2aJ*. 

10. 

—  7  7a'" +2  by  11a? 

58         TO  DIVIDE  A   POLYNOMIAL   BY  A  MONOMIAL. 

47.  To  Divide  a  Polynomial  by  a  Monomial. 

Since         (a  —  b)c  =  ac  —  be ; 

therefore        — — —  =  a  —  b. 
c 

Also  since  (a  —  b)   x  (  —  c)  =  — «c  +  6c; 

,,   -           —  ac  -f-  be  , 

therefore         ! =  a  —  b. 

—c 

Also  since  (a  —  b  -\-  c)  ab  =  a25  —  ab2  +  a&c ; 

,,       £  a2b  —  ab2  +  abc  7    . 

therefore      ! =  a  —  b  -f-  c. 

a& 

Hence  we  have  the  following  rule  : 

To  divide  a  •polynomial  by  a  monomial,  divide  each  term  of 

the  dividend  separately  by  the  divisor. 

For  example 

(8a3  -  6a26  +  2a2c)  -*-  2a2  =  4a  -  36  +  c. 

(9a  -  \2y  -f  3z)  -:-  -3  =  -3x  +  4y  -  z. 

(36a362  -  24a265  -  20a462)  -s-  4a26  =  9ab  -  Qb4  -  5a2b. 


(2x2  —  bxy  —  fa;'2?/3)  -s- 

~ix  = 

-4a;  +  lOy  +  3a;*/3. 

EXAMPLES. 

Divide 

1.   -35a6  by  7a;3. 

Ans.  —5a;3. 

2.    x3y3  by  x2y. 

xy2. 

3.   4aW  by  a&2c2. 

Aac, 

4.    x2  —  2xy  by  x. 

x  -  2y, 

5.    a6  —  7xb  +  4a;4  by  a;2. 

a;4 

-  7.r3  +  4a;2. 

6.    10a/7  -  8a;6  +  3a;4  by  a;8. 

10a;4 

—  8a;3  +  3a;. 

7.    15a;5  -  25a;4  by  -  5a;8. 

—3a;2  -f-  ox. 

8.    27a;6  -  36a;5  by  9a;5. 

Sx  -  4. 

9.    -24a;6  -  32a;4  by  -8x\ 

3a;3  +  4.r. 

10.    a2  —  ab  —  ac  by  —a. 

— a  -f-  b  -+-  c. 

48.  To  Divide  one  Polynomial  by  Another.  —  Let 

it  be  required  to  divide  a3  -f-  2aa  —  3a  by  a2  +  3a. 

Here  we  are  to  find  a  quantity  which  when  multiplied  by 
the  divisor  will  produce  the  dividend.      Hence  the  dividend 


TO   DIVIDE    ONE   POLYNOMIAL    BY  ANOTHER.  59 

is  composed  of  all  the  partial  products  arising  from  the 
multiplication  of  the  divisor  Ixy  each  term  of  the  quotient 
(Art.  39).  Arranging  both  the  dividend  and  divisor  accord- 
ing to  descending  powers  of  a,  we  see  that  the  first  term  a3 
of  the  dividend  is  the  product  of  the  first  term  a2  of  the 
divisor  by  the  first  term  of  the  quotient  (Art.  43) ;  therefore, 
dividing  a3  by  a2  we  obtain  a  for  the  first  term  of  the 
quotient.  Multiplying  the  whole  divisor  by  a  we  obtain 
a*  +  3a2  for  the  partial  product  of  the  divisor  by  the  first 
term  of  the  quotient ;  subtracting  this  product  from  the 
dividend  we  obtain  the  first  remainder  —a2  —  3a,  which  is 
the  product  of  the  divisor  by  the  remaining  terms  of  the 
quotient,  and  consequently  the  first  term  —  a2  of  this  product 
is  the  product  of  the  first  term  of  the  divisor  by  the  second 
term  of  the  quotient.  Dividing  therefore  this  first  term  —a2 
by  the  first  term  of  the  divisor  a2  we  obtain  —1  for  the 
second  term  of  the  quotient.  Multiplying  the  whole  divisor 
by  —1  we  obtain  —a2  —  3a  for  the  product  of  the  divisor  by 
the  second  term  of  the  quotient ;  subtracting  this  product 
there  is  no  remainder.  As  all  the  terms  in  the  dividend 
have  been  brought  down,  the  operation  is  completed.  Hence 
a  —  1  is  the  exact  quotient. 

The  work  may  be  arranged  as  follows : 

Divisor.  Dividend.  Quotient. 

+  3a)  a3  +  2a2  —  3a  (a  —  1 
a3  +  3a2 


a2  -  3a 
a2  —  3a 


It  will  be  observed  that  in  getting  each  term  of  the 
quotient  in  this  example,  we  divide  that  term  of  the  divi- 
dend containing  the  highest  power  of  a  by  the  term  of  the 
divisor  containing  the  highest  power  of  the  same  letter , 
and  therefore,  when  the  dividend  and  divisor  are  arranged 
according  to  descending  powers  of  a,  any  term  of  the 
quotient  is  found  by  dividing  the  first  term  of  the  divisor 


GO  TO   DIVIDE    ONE  POLYNOMIAL   BY  ANOTHER. 

into  the  first  term  of  the  dividend,  or  into  the  first  term  of 
one  of  the  remainders. 

Hence  for  the  division  of  one  polynomial  by  another,  we 
have  the  following 

Rule. 

Arrange  both  dividend  and  divisor  according  to  ascendinrj 
or  descending  powers  of  some  common  letter. 

Divide  the  first  term  of  the  dividend  by  the  first  term  of 
the  divisor,  and  write  the  result  for  the  first  term  of  the 
quotient ;'  midtiply  the  whole  divisor  by  this  term,  subtract 
the  product  from  the  dividend,  and.  to  the  remainder  join  as 
many  terms  from  the  dividend,  taken  in  order,  as  are  required. 

Divide  the  first  term  of  the  remainder  by  the  first  term  of 
the  divisor,  and  write  the  result  for  the  second  term  of  the 
quotient;  multiply  the  vohole  divisor  by  this  term,  and  subtract 
the  product  from  the  last  remainder. 

Continue  this  operation  until  the  remainder  becomes  zero, 
or  until  the  first  term  of  the  remainder  will  not  contain  the 
first  term  of  the  divisor. 

This  method  of  dividing  is  similar  to  long  division  in 
Arithmetic,  i.e.,  we  break  up  the  dividend  into  parts,  and 
find  how  often  the  divisor  is  contained  in  each  part ;  and  then 
the  sum  of  these  partial  quotients  is  the  complete  quotient. 
Thus,  in  the  example  just  solved,  a3  -f-  2a2  —  3a  is  divided 
by  the  above  process  into  two  parts,  viz.,  a3  4-  3a2,  and 
—a2  —  3a,  and  each  of  these  is  divided  by  a2  +  3a,  giving 
for  the  partial  quotients  a  and  —  1  ;  thus  we  obtain  the 
complete  quotient  a  —  1. 

Note.  —  The  divisor  is  often  put  on  the  right  of  the  dividend  and 
the  quotient  beneath  the  divisor  as  follows: 

Dividend.  Divisor. 

a2  +  2ab  +  b2  \a  -f  b 
a2  +    ab  a  +  b  Quotient. 

ab  +  b2 
ab  +  b2 


EXAMPLES.  61 

It  is  of  great  importance  to  arrange  both  dividend  and 
divisor  according  to  ascending  or  descending  powers  of  some 
common  letter ;  and  to  attend  to  this  order  in  every  part  of  the 
operation. 

EXAMPLES. 

1.  Divide  24a;2  —  S5xy  +  2\y2  by  Sx  —  Sy. 
The  operation  is  conveniently  arranged  as  fellows : 

8a;  —  oy)  24a;2  —  65xy  +  2\y2(ox  —  7y. 
24a2  -     9xy 

—  06.17  -f  2hf 
-56xy  +  21/ 

Divide 

2.  x2  +  3.?;  +  2  by  x  -f  1.  ^4?is.  a;  +  2. 

3.  x2  -  7x  +  12  by  a:  -  3.  x  —  4. 

4.  a;2  -  11a  -f  30  by  a;  -  5.  a?  —  6. 

5.  x2  -  49a;  4-  GOO  by  x  -  25.  x  -  24. 

6.  3a;2  +  10a;  +  3  by  x  +  3.  3a;  +  1. 

7.  2a;2  +  11a;  -f  5  by  2x  +  1.  x  +  5. 

8.  hx2  +  Ux  +  2  by  x  +  2.  5a;  +  1. 

9.  2a2  +  17a  +  21  by  2a;  +  3.  x  +  7. 

10.  5a2  +  16a:  +  3  by  a;  +  3.  5a;  +  1. 

11.  Divide  3a4  -  10a3/a  +  22a262  -  22a63  +  1564  by 
a2  -  2ab  +  362. 

The  operation  is  written  as  follows. 
oa  -  2ofi  +  362)3a4  -  10rt36  +  22a%2  -  22ab*  +  15£4(3a2  -  4a&  +  5b- 
3a4-    G«%  +    9a262 


-  4«%+  13a%»-22aP 

-  4a86+    S«262-12«&3 

5a262  -  10a6»  +  156* 
5a262  -  10a£3  +  1564 

12.    Divide  x'J—5x5+7xs+2x2-Gx  —  2  by  1  +  2a;—  3a;2 -f  x\ 
Arrange  both  dividend  and  divisor  according  to  descending 
powers  of  x,  and  arrange  the  work  as  follows : 


02                                                 EXAMPLES. 

xi  _  5^5              _j_  7^3  +  2x2  -  6aj  -  2  |a;4  - 

-  3a;2  +  2x  +  1 

x7  -  3a;5  +  2x4  +    x3                                 x3  - 

-  2a;   -  2 

-  2a5  -  2a4  +  6a;3  +  2x2  -  6x 

-  2a;5               +  6x3  -  4a;2  -  2x 

—  2a;4              +  Qx2  —  4x  —  2 

-  2a;4              +  6a;2  -  4x  -  2 

We  might  have  arranged  the  dividend  and  divisor  accord- 
ing to  ascending  powers  of  x  as  follows : 
_2  -  6a;  +  2a;2  +  7a;3  -  5a;5  +  a;7 1 1  +  2a;  -  3a;2  +  a;4 

_2  -  4a;  -f  Gx2  -  2a;4  -2  -  2x  +  a;3 


—  2a;  —  4a;2  +  7a;3  +  2a;4  —  5a;5 

-  2'a;  -  4ar  +  6a;3  -  2a;5 

a;3  -f-  2a;4  —  3a;5  +  x1 
xs  +  2a;4  —  3a;5  -f  x1 

We  thus  obtain  the  same  quotient  we  had  before,  though 
the  terms  are  in  a  different  order. 

13.   Divide  a3  +  b3  +  c3  —  oabc  by  a  +  b  -f-  c. 
Arrange  the  dividend  according  to  descending  powers  of  a 
a3  -  Sabc  -f     b3  -f  c3\a    +     b  -f    c 


a3  +     «2fr  -f-  a2c 

a2  - 

Sabc 
abc 

ab  —  ac  -+-  b2  —  be  +  c2 

—  a2b  —  a2c  — 

-  a2b  -  ab2  - 

-  a2c  + 

—  a2c 

ab2  - 

2abc 
abc  —  ac2 

ab2  - 
ab2 

abc  +  ac2  -f  &3 

+  b3    +  62c 

— 

a  6c  -f-  ac2  —  b2c 

abc             —  b2c  —  be2 

ac2  +  be2  -\-    c3 
ac2  +  be2  +    c8 

EXAMPLES.  63 

In  this  example  we  arrange  the  terms  according  to  de- 
scending powers  of  a  ;  then  when  there  arc  two  terms,  such 
as  a2b  and  crc,  which  involve  the  same  power  of  a,  we 
select  a  new  letter,  as  6,  and  put  the  term  which  contains  b 
before  the  term  which  does  not;  and  again  of  the  terms  ab~ 
and  «6c,  we  put  the  former  first  as  involving  the  higher 
power  of  b. 

14.    Divide  a:4  +  4a4  by  x2  +  2ax  +  2a'2. 

x*  +  4  a4  \x2  +  %ax  +  2a2 

x4  +  2GUE8  4-  2aV     jc2  —  2aa;  4-  2a2 

-  2ax*  -  2aV 

—  2«a-3  —  Aa2x2  —  4a8se 


2a2a;2  +  4a8»  +  4a4 
2aV  +  4a8a;  +  4a4 
Divide 

15.  k8  -  5a;4  +  9:c3  -  Oa;2  -  aj  +  2  by  a;2  -  Sx  +  2. 

-4ns.  jc3  -  2x2  +  x  +  1. 

16.  x5  -  2x4  -  4a;3  +  19a;2  -  31a;  4-  15  by  a;3  -  7a;  4-  5. 

-4ws.  x2  -  2a;  4-  3. 

17.  a3  4-  &3  4-  3a6c  -  c3  by  o  +  6-c. 

Ans.  a2  —  aft  4-  ac  4-  ^  +  &c  +  c2. 

18.  a;4  4-  64  by  x2  4-  4a;  4-  8.  x2  -  \x  4-  8. 

19.  a6  -  66  by  a8  -  2tt26  4-  2«62  -  63. 

^7?s.  a8  4-  2d*b  4-  2a62  4-  &3. 
When  the  coefficients  are  fractional  we  may  still  use  the 
ordinary  process,  combining  the    fractional   coefficients  by 
the  rules  of  Arithmetic. 

20.  Divide  Jx8  4-  ?W  +  tW  b}T  ix  +  to- 

la;3  4-  TW  +  A?/3  [fo  +  to 

Ix3  +  Jafy  £ar  -  \xy  4-  i*/2 


64  DIVISION    WITH   THE   AID    OF  PARENTHESES. 


Divide 
21. 


K  - 

fci2z 

+ 

^-ax2 

—  27a;3  by  \a 
Ans. 

—  3». 

K  - 

3aa;  +  9#2. 

1  n*  - 

15  T 

"  tW 

!  + 

A«  ~ 

A  by  la  -  J 

■   K 

—  -$a  -+-  ttt- 

22. 

49.  Division  with  the  Aid  of  Parentheses.  —  Some- 
times it  is  found  convenient  to  divide  with  the  aid  of  paren- 
theses thus : 

1 .  Divide  a;3—  (a+b+c)x2+  (ab+ac+bc)x—abc  by  x—c. 
x3  —  (a  -+-  b  +  c)x2+  (ab  -f  ac  -f-  bc)x  —  abc  \x  —  c 
xs  -  ex2  x2—(a+b)x+ab 


—  (a  +  b)x2  -f  (a6  +  ac  +  6c) a; 

—  (a  -f  6)a;2  -f-  («  +  b)cx 

abx  —  «6c 
abx  —  abc 
Divide 

2 .  a2x* +  ( 2ac — b2) x2 -f- c2  by  ax2 — bx -f- c.    Ans.  ax2 +bx-\-c. 

3.  a;3  -f-  (a  -f-  6  +  c)ar  -f-  (ab  -f  ac  -+-  bc)x  -f-  a&c  by  a;  -f-  6. 

Jbs.  a;2  -f-  (a  +  c).y  -+-  «c. 

4.  aa;3  —  (a2  -f  6) a;2  +  o2  by  ax  —  b.  x2  —  ax  —  b. 
b.  a2(b  +  c)  +  b2(a  -  c)  +  c2(<x  -  6)  +  a&c  by  a  +  b  +  <?. 

^l?is.  a  (6  -f-  c)  —  6c. 
50.  Where  the  Division  cannot  be  Exactly  Per- 
formed. —  In  the  examples  given  thus  far  the  divisor  has 
been  exactly  contained  in  the  dividend.     It  may  happen,  as 
in  Arithmetic,  that  the  division  cannot  be  exactly  performed. 
In  such  cases  it  is  a  good  exercise  for  the  student  to  deter- 
•nine  the  accurate  value  of  the  remainder. 
Divide  a;3  —  Co;"2  +  11a;  +  2  by  x  -  2. 
a;3  -  Gar  +  11a;  +  2\x   -  2 


Xs  -  2x2                        x2 

-  4a;  +  3 

—  Ax2  +  11a; 

—  Ax2  +    8a; 

3a;  -f  2 

3a;  -  G 

IMPORTANT   EXAMPLES   IN   DIVISION. 


05 


The  division  can  be  carried  no  further  without  fractions, 
because  x  will  not  go  into  8.  We  therefore  express  the 
result  in  the  same  way  as  in  Arithmetic,  that  is,  by  adding 
to  the  quotient  a  fraction  of  which  the  numerator  is  the 
remainder  and  the  denominator  the  divisor.  Thus  the  result 
is 


Cs2  +  lis  +  2 
x  —  2 


=  a," 


4x  +  3  + 


x  —  2 


EXAMPLES. 


Find  the  remainder  when 


x*  —  (jx2  +  123—17  is  divided  by  x  —  3. 
3x3  —  7x  —  9  is  divided  by  x  +  1. 
2x*  -\-  bx2  —  4x  —  7  is  divided  by  x  +  2. 
4a3  +  7x2  —  3x  —  33  is  divided  by  4x  —  5. 
27a3  +  9x2  —  3x  —  5  is  divided  by  3a;  —  2. 
16s3  -  19  +  39s  -  46x2  is  divided  by  8x  -  3. 
8x  —  8x2  +  5x3  -\-  7  is  divided  by  5x  —  3. 


.4?is.  —8. 

—  5. 

5. 

-18. 

5. 

-10. 

10. 


51.  Important  Examples  in  Division.  —  The  follow- 
ing examples  are  very  important ;  they  may  be  easily  verified, 
and  should  be  carefully  noticed. 


I. 


t 


=  *  +  y, 


—  =  x2  +  xy    +  ?/2, 

x 


x  —  y 

t 

II 


X9 


t   _ 


xs  +  x2y  +  xy2  4-  2/3, 


x  —  y 
and  so  on ;  the  terms  in  the  quotient  all  beiug  positive. 


II. 


ft 


x  +  y 

x*  -  y*  =  xs  _ 

x  +  y 

x*  —  if 


._»  +  y 


-  y, 

x2y  +  xy2 
xh  —  x4y  +  xPy2 


xhf  +  «2/4  -  2/B, 


QQ  IMPORTANT  EXAMPLES  IN  DIVISION. 

and  so  on  ;    the  terms  in  the  quotient  being  alternately  posi- 
tive and  negative. 

x2  -  xy  +  y2, 


x3  -f  ?y3 

x  +  y 

x5  -f  yb 

x  +  y 

x1  4-  if 

II.     •"     '    *    =  a4  -  afy  4-  x2y2  -  xy3  4-  r/4, 

/v,7      i      „,7 

=  a;6  —  xhj  4-  a;4?/2  —  x3y3  4-  x*2?/4  —  &?/5  -f-  ?/6, 
a  +  2/ 

and   so   on ;    the   terms   in   the   quotient    being    alternately 
'positive  and  negative. 

The  student  can  verify  these  results  in  any  particular  case, 
and  carry  on  these  operations  as  far  as  he  pleases,  and  he 
will  thus  gain  confidence  in  the  truth  of  the  following  state- 
ments for  which  we  shall  hereafter  give  a  general  proof.  See 
College  Algebra,  Art.  170. 

These  different  cases  may  be  conveniently  arranged  in  the 
following  concise  statements : 

xn  —  yn  is  divisible  by  x  — -  y  if  n  be  any  whole  number. 
(1)  That  is:  the  difference  of  any  tivo  equal  poivers  of  two 
numbers  is  always  divisible  by  the  difference  of  the  two 
numbers. 

xn  _  yn  js  divisible  by  x  +  y  if  n  be  any  even  whole 
number.  (2)  That  is  :  the  difference  of  any  two  equal  even 
powers  of  two  numbers  is  always  divisible  by  the  sum  of  the 
numbers. 

xn  _|_  yn  is  divisible  by  x  4-  y  if  n  be  any  odd  whole 
number.  (3)  That  is  :  the  sum  of  any  two  equal  odd  power* 
of  two  numbers  is  always  divisible  by  the  sum  of  the  numbers. 

x"  4-  yn  is  never  divisible  by  x  +  y  or  x  —  y,  when  n  i* 
an  even  whole  number. 

EXAMPLES. 

Write  the  results  in  the  following  by  these  three  state- 
ments. 

1.  a5  —  b5  -T-  a  —  b.     Ans.  a4  4-  a*b  4-  a~b~  4-  ab3  4-  bA. 

2.  X3  —  1   -r-  x  —   1.  x*  +  X  +   1. 


a2  _  4a  +  16. 

-x2  +  3xy  +  Ay2. 

2x3y3  -\-  4x2y  —  Sy2. 

x2  —  3xy  +  Ay2. 

5a2b'2  +  ab  —  4. 

2a  —  36  +  4c. 

_3sc2+2?/2-f  4xy. 

|a  _  ib  _  c# 

a;  —  6. 

EXAMPLES.  67 

3.  8a3  -  b3  --  2a  -  6.  Ans.  4a2  +  2r/6  +  b2. 

4.  8a;3  -  27?/3  -=-  2sb  -  3y.  4or  +  Q>xy  +  9?/2. 

5.  x4  -  16?/4  --  a  +  2y.  Xs  -  2x2y  -f-  Axy2  -  8y3. 

6.  16.v4  -  ?/4  -f-  2sb  +  ?/.  8a;3  —  4a%  +  2xy2  —  y3. 

7.  x3  +  1  --  sb  +1.  a2  -  SB  +  1. 

8.  a3  +  64  -r-  a  +  4. 
Divide 

9.  3SB8  -  9a;2?/  -  12a;?/2  by  -3<b. 
(lOj)  4a;4?/4  -  8sBy  +  6xy3  by  -  2sb#.      - 

11.  x3y  —  3x2y2  +  4a;#3  by  xy. 

12.  -15a3//  -  3a262  -f  12a6  by  -3a6. 

13.  —3a2  +  lab  —  6ac  by  —fa. 

14.  f.rV2  -  3a;3?/4  -  GzV  by  -fa;3?/2. 

15.  \a2x  —  -^abx  —  facsB  by  |osb. 

16.  x2  —  IIsb  +  30  by  sb  —  5. 

17.  sc2  -  7sb  +  12  by  sb  -  3.  a;  ~  4. 

18.  3a;2  +  x  -  14  by  sb  -  2.  3jb  +  7. 

19.  Go;2  -  31sb  +  35  by  2a;  -  7.  3a  -  5. 

20.  15a2  +  17aa;  —  4a;2  by  3a  -f  Ax.  5a  —  x. 

21.  60a;2  —  4xy  —  A5y2  by  10a;  —  9y.  6x  +  5?/. 

22.  —  Axy  —  loy2  +  96a;2  by  12jb  —  by.  8sb  -f  3?/. 
(5^>  100a;3  -  3sb  -  13a;2  by  3  +  25sb.                       4.x2  -  sb. 

24.  7a;3  +  96a;2  —  28a;  by  7a;  —  2.  x2  +  14a;. 

25.  x5-{-x4y— x3y2-\-x3— 2xy2+y3byx2+xy  — y2.     x3+x—y. 

26.  2a;3-  8sb  +  xA+  12  -  7a;2 by  se2+  2  -  3a;.       x2  +  ox  +  6. 

27.  8sb8-4jb2-  128a;  +  a;4-192bya;2-16.       sb2+8jb  +  12. 

28.  a;9-?/9by  a;2+  a-?/+2/2.       «7-  sB^+sBY-sBY+SBt/6-?/7. 

29.  2a4+27a63-8164bya+36.      2a8-6a26+18a62-2763. 

30.  x5+  x*y+  x3y2+  xY+xif+ifhy  x3+y3.       x2+xy  +y2. 

31.  |a2c3  +  yf^a5  by  ia2  +  -|ac.  ^a3  -  fa2c  +  |ac2. 

32.  T^a4-|a3-Ia2+|a+^by|a2-f-a.       §a2-Ja-f. 

33.  sb6— 1  by  sb  — 1.  (Art.  51.)        sB5+SB4+sB8+aj2+JB+l. 

34.  a;4  -  81?/4  by  sb  -  Sy.  x3  +  3a;2?/  +  9a;?/2  +  27ys. 

35.  a;5  —  ?/5  by  x  —  ?/.  a;4  +  x3y  +  sb2^2  +  av/3  +  ?/4. 

36.  a9  -  b»  by  a3  -  b3.  a6  +  «363  +  &6- 

37.  27JB8  +  8/  by  3a;  +  2y.  9x2  -  6xy  +  4/. 


68     EQUATION   OF   CONDITION  —  UNKNOWN    QUANTITY, 


.    CHAPTER    VI. 

SIMPLE    EQUATIONS    OF    ONE     UNKNOWN 
QUANTITY. 

52.  Equations  —  Identical  Equations.  —  An  Equation 
is  a  statement  in  Algebraic  language  that  two  expressions 
are  equal.     Thus,        2x  +  A  =  x  +  8 

is  an  equation  ;  it  states  that  the  expression  2x  -f-  4  is  equal 
to  the  expression  x  -f-  8. 

The  two  equal  expressions  thus  connected  are  called  sides 
or  members  of  the  equation.  The  expression  to  the  left 
of  the  sign  of  equality  is  called  the  first  side  or  member, 
and  the  expression  to  the  right  is  called  the  second  side  or 
member.     Every  equation  has  two  members. 

An  Identical  Equation,  or  briefly  an  Identity,  is  one  in 
which  the  two  members  are  equal  whatever  numbers  the 
letters  represent.  Thus,  the  following  are  identical  equa- 
tions :  n  n 

x  +  3+x  +  4:  =  2x+7, 

(a  -f  x)  (a  —  x)  —  a2  +  x2  =  0 ; 

that  is,  these  Algebraic  statements  are  necessarily  true, 
whatever  values  we  assign  to  x  and  a.  All  the  equations 
used  in  the  previous  chapters  to  express  the  relations  of 
Algebraic  quantities  are  identical  equations,  because  they 
are  true  for  all  values  of  these  quantities. 

53.  Equation  of  Condition  —  Unknown  Quantity. 
—  An  Equation  of  Condition  is  one  which  is  true  only  when 
the  letters  represent  some  particular  value.  For  example, 
the  equation,  a  +  7  =  12, 

cannot  be  true  unless  x  =  5,  and  is  therefore  an  equation  of 
condition. 


AXIOMS.  69 

An  equation  of  condition  is  called  briefly  an  equation. 

The  letter  whose  value,  or  values,  it  is  required  to  find 
is  called  the  unknown  quantity.  Thus  x  is  the  unknown 
quantity  in  the  above  equation. 

To  solve  an  equation  means  to  find  the  value,  or  values, 
of  the  unknown  quantity  for  which  he  equation  is  true. 
These  values  of  the  unknown  quantity  are  said  to  satisfy  the 
equation,  and  are  called  the  roots  of  the  equation. 

An  equation  which  contains  only  one  unknown  quantity  is 
called  a  simple  equation,  or  an  equation  of  the  first  degree, 
when  the  unknown  quantity  occurs  only  in  the  first  power. 
It  is  usual  to  denote  the  unknown  quantity  by  the  letter  x. 
The  equation  is  said  to  be  of  the  second  degree  or  a  quadratic 
equation  when  x2  is  the  highest  power  of  x  which  occurs,  and 
so  on.* 

Thus  2x  +  G  =  x  +  8, 

and  ax  -f  b  =  c 

are  simple  equations. 

x2  -  2x  =  3 
is  a  quadratic  equation. 

54.  Axioms.  —  An  Axiom  is  a  self-evident  truth.  The 
operations  employed  in  solving  equations  are  founded  upon 
the  following  axioms  : 

1.  If  equal  quantities  be  added  to  equal  quantities,  the 
sums  will  be  equal. 

2.  If  equal  quantities  be  taken  from  equal  quantities,  the 
remainders  will  be  equal. 

3.  If  equal  quantities  be  multiplied  by  equal  quantities, 
the  products  will  be  equal. 

4.  If  equal  quantities  be  divided  by  equal  quantities,  the 
quotients  will  be  equal. f 


*  The  equation  is  supposed  to  be  reduced  to  such  a  form  that  the  unknown 
quantity  is  found  only  in  the  numerators  of  the  terms,  and  that  the  exponents  of  ita 
powers  are  expressed  by  positive  integers. 

t  If  the  divisors  are  different  from  zero. 


70  CLEARING   OF  FRACTIONS. 

5.  Like  powers  and  like  roots  of  equal  quantities  are 
equal. 

These  axioms  may  be  summed  up  in  the  following  one : 
If  the  same  operations  be  performed  on  equal  quantities,  the 
results  will  be  equal. 

In  the  solution  of  equations  there  are  two  operations 
of  frequent  use.  These  are  (1)  clearing  the  equation  of 
fractions,  and  (2)  transposing  the  terms  from  one  member 
to  the  other  so  that  the  unknown  quantity  shall  finally  stand 
alone  as  one  member  of  the  equation. 

55.  Clearing  of  Fractions.  —  Consider  the  equation 

2!  4_  £?  _i_  E  —  2. 
2       3       6 

Multiplying  each  term  by  2  x  3  x  6  (Axiom  3),  we  get 
3  X  6a;  +  2  x     6a;  +      2  x  Sx  =  2  x  3  X  6  X  2, 
or  18a;  -f-  12a;  +  6a;  =  72; 

dividing  each  term  by  6  (Axiom  4),  we  get 

3a;  +  2x  +  x  =  12, 
or  6x  =  12. 

Instead  of  multiplying  each  term  by  2  x  3  x  6,  we  might 
multiply  each  term  by  the   least  common    multiple   of   the 
denominators,  which  is  6,  and  get  immediately 
3x  -f  2a;  +  x  =  12. 
Hence   to   clear  an   equation   of   fractions,  we   have   the 
following 

Eule. 

Multiply  each  term  of  the  equation  by  the  least  common 
multiple  of  the  denominators. 

Clear  the  following  equation  of  fractions : 

3       5  "  7  "~ 
Here  the  least  common  multiple  of   the  denominators  is 
the  product  of  the  denominators,  3,  f>,  and  7.     Multiplying 
each  term  by  it,  we  get 

35a;  +  21a;  -  15a;  =  315, 
or  41a;  =  315. 


TRANSPOSITION.  71 

Clear  the  following  equation  of  fractions  : 

4       6       8       12 

Here  24  is  the  least  common  multiple  of  the  denominators. 
Multiplying  each  term  by  it,  we  get 

6a;  +  4a;  +  3x  +  2a;  =  96, 
or  15a;  =  96. 

56.  Transposition.  —  To  transpose  a  term  is  to  change 
it  from  one  member  of   an   equation  to  the   other  without 
destroying  the  equality  of  the  members. 
Suppose,  for  example,  that  x  —  a  =  b. 
Add  a  to  each  member  (Axiom  1)  ;  then  we  have 
x  —  a  -f  a  =  b  -f  a ; 
therefore,  since  —a  and  +a  cancel  each  other,  we  have 
x  =  b  +  a. 
Again,  suppose  that  x  -f-  b  =  a. 

Subtract  b  from  each  member  (Axiom  2)  ;  then  we  have 
x  +  b  —  b  =  a  —  b\ 
therefore,  since  +b  and  —b  cancel  each  other,  we  have 
x  =  a  —  b. 
Here  we  see,  in  these  two  examples,  that  —a  has  been 
removed  from  one  member  of   the  equation,   and   appears 
as  +a  in  the  other;  and  -\-b  has  been  removed  from  one 
member  and  appears  as  —b  in  the  other. 

It  is  evident  that  similar  steps  may  be  employed  in  all 
cases.     Hence  we  have  the  following 

Rule. 

Any  term  may  be  transposed  from   one   member  of  an 
equation  to  the  other  by  changing  its  sign. 

It  follows  from  this  that  the  sign  of  every  term  of  an  equa- 
tion may  be  changed;  for  this  is  equivalent  to  transposing 
every  term,  and  then  making  the  first  and  second  members 
change  places.     Thus,  for  example,  suppose  that 
Ax  -  8  =  2a;  -  16. 


72  SOLUTION   OF  SIMPLE   EQUATIONS. 

Transposing  every  term,  we  have 

-2x  +  16  =  -4a;  +  8, 
or  -4x  +     8  =  -2x  +  16, 

which  is  the  original  equation  with  the  sign  of  every  term 
changed. 

This  result  can  also  be  obtained  by  multiplying  each  term 
of  the  original  equation  by  —  1  (Axiom  3) . 

57.  Solution  of  Simple  Equations  with  One  Un- 
known Quantity.  —  Find  the  value  of  x  in  the  equation 

x  _  18  _  x  _  x 
2        5   ~  4       5' 

The  least  common  multiple  of  the  denominators  is  20. 
Multiplying  each  term  by  20,  we  get 

10&  —  72  =  hx  —  4x. 

Transposing  the  unknown  terms  to  the  first  member,  and 
the  known  terms  to  the  second,  we  have 

10a;  -  5x  -f-  4x  =  72. 

Collecting  the  terms,  we  have 

9x  =  72. 

Dividing  each  member  by  9  (Axiom  4),  we  have 

x  =  8. 

We  can  now  give  a  general  rule  for  solving  any  simple 
equation  with  one  unknown  quantity. 

Rule. 

Clear  the  equation  of  fractions,  if  necessary;  transpose  all 
the  terms  containing  the  unknown  quantity  to  the  first  member 
of  the  equation,  and  the  known  quantities  to  the  secona:,  and 
collect  the  terms  of  each  member.  Divide  both  members  by 
the  coefficient  of  the  unknown  quantity,  and  the  second  member 
is  the  value  required. 


EXAMPLES.  73 


EXAMPLES, 


1.  Solve  9x  +  35  =  75  +  5x. 

Here  there  are  no  fractions  ;  transposing,  we  have 
9a;  —  5.v;  =75  —  35. 
Collecting  terms,  Ax  =  40. 

Dividing  by  4,  x  =  10. 

It  is  very  important  for  the  student  to  acquire  the  habit 
of  occasionally  verifying,  that  is,  proving  the  truth  of  his 
results.  The  habit  of  applying  such  proofs  tends  to  con- 
vince the  student,  and  to  make  him  self-reliant  and  confident 
in  his  own  accuracy. 

To  verify  the  result,  in  the  case  of  simple  equations,  we 
substitute  the  value  of  the  unknown  quantity  in  the  original 
equation  ;  if  the  two  members  are  equal  the  result  is  said  to 
be  verified,  or  the  equation  satisfied. 

Thus,  in  the  last  example,  10  is  the  root  of  the  proposed 
equation  (Art.  53).  We  may  verify  this,  i.e.,  we  may  show 
that  x  =  10  satisfies  the  original  equation  by  putting  10  for 
x  in  that  equation. 

Thus  9  x  10  -f-  35  =     75  +     5  x  10, 

or  90  +  35  =     75  +  50, 

or  125  =  125, 

which  is  clearly  true.     Hence,  since  the  two  members  are 
equal,  x  =  10  satisfies  the  equation. 

2.  Solve  5(<b  -  3)  -  7(6  -  x)  +  3  ==  24  -  3(8  -  x). 
Removing  parentheses, 

5X  _  15  _  42  +  7x  +  3  =  24  -  24  +  Sx. 
Transposing,     ox  +  7as  —  3a;  =24  —  24  +  15+42  —  3. 
Collecting  terms,  9x  =  54. 

Dividing  by  9,  x  =  6. 

We  may  verify  this  result  by  putting  6  for  x  in  the  given 
equation. 

Thus  5(6  -  3)  -  7(6  -  6)  +  3  =  24  -  3(8  -  6), 
or  15  -f  3  =  24  -  6, 

or  18  =  18. 


74  EXAMPLES. 

3.    Solve  5x  -  (4a:  -  7)  (3.x-  -  5)  =  6  -  3(4a;  -  9)  (a;  -  1). 

Performing  the  multiplications  indicated,  we  have 
bx  -  (12a;2  -  41a;  +  35)  =  6  -  3  (4a/2  -  13a;  +  9). 
Removing  the  parentheses,  we  have 

5x  -  12a;2  +  41s  -  35  =  6  -  12a;2  +  39a;  -  27. 
Erasing  the  term  — 12a;2  on  each  side,  and  transposing,  we 
have  bx  +  41a?  -  39a;  =     6-27  +  35. 

Collecting  terms  7a;  =  14. 

.•.     x  =     2. 
We  may  verify  this  result  by  putting  2  for  x  in  the  given 
equation. 

The  first  member  becomes 

10  -  (8  -  7)  (6  -  5)  =  10  -  1  =  9  ; 
and  the  second  member  becomes 

6  -  3(8  -  9)(2  -  1)  =  6  -  3(-l)  =  9. 
Thus,  since  these  two  results  are  the  same,  x  =  2  satisfies" 
the  equation. 

Note.  —  In  the  first  line  of  the  solution  of  Ex.  3,  we  did  not 
remove  the  parentheses  until  we  performed  the  multiplications.  The 
beginner  is  recommended  to  put  down  all  his  work  as  full  as  we  have 
done  in  this  example,  in  order  to  insure  accuracy. 

4.  Solve  8a;  -  5[x  -  \G  -  5(x  -  3)  J]  =  4a;  +  1. 
Removing  parentheses,  we  have 

8a;  -  5[a;  -  21  +  5a?]  =  Ax  -f-  1, 
8a;  —  5x  -f  105  —  25a;  =  4a;  -f-  1. 
.-.     -26a;  =  -104. 
x  =  4. 

Solve  the  following  equations  : 

5.  2a;  -f  7  =  3a;  -f  3.  Ans.  4. 

6.  24a;  -  49  =  19a;  -  14.  7. 

7.  lGx  —  11  =  7a;  +  70.  9. 

8.  8(x  -  1)  +  170  -  3)  =  4(4a;  -  9)  -f  4.  3. 

9.  5a?  -  6 (re  -  5)  =  2 (a;  +  5)  +  5(x  -  4).  5. 

10.  S(x  -  3)  -  (6  -  2a;)  =  2 (a;  +  2)  -  5(5  -  x).       3. 

11.  3(169  -  x)  -  (78  +  a;)  =  29a;.  13. 

12.  7a;  -  39  -  10a;  +  15  =  100  -  33a;  +  26.  5. 


FRA  CTIOXA  L    E  Q  l\  1 TIONS  —  EXA  MTL  ES.  75 

13.  118  —  65a;  —  123  =  15a;  +  35  —  120x.  Ans.  1. 

14.  157  -  21  (x  +  3)  =  163  -  l5(2a;  -  I).  1G. 

15.  97  -  5(.r  +  20)  =  111  -  S(x  +  3).  30. 
10.  x  _  [3  +  J.r  -  (3  +  a;)|]  =  5.  5. 

17.  14a;  -  (5a;  —  9)  —  |4  —  3a;  -  (2a;  -  3)  \  =  30.       2. 

18.  5x  -  (Sx  —  7)  —  J4  —  2a;  -  (6a;  -  3)  \  =  10.       1. 
58.    Fractional  Equations.  —  The  following  are  some 

of  the  most  useful  methods  of  solving  fractional  equations. 

EXAMPLES. 

,     o  i      5a;  -f  4       7x  +  5       .3       x  —  1 

1.    Solve ! —  =  o- — . 

2  10  5  2 

5-|  =  -2^  ;  the  least  common  multiple  of  the  denominators 

is  10;  multiplying  by  10,  we  have 

5(5./'  +  4)  —  (7a;  +  5)  =  06  —  o(x  —  1) ; 

removing  parentheses, 

25a;  -f-  20  —  7x  —  5  =  56  —  5.7;  -f  5 ; 

transposing,  2ox  —  7x  +  ox  =  56  +  5  —  20  +  5; 

collecting  terms,  23.tj  =  4G  ; 

.-.     x  =     2. 

Xote.  —  Mistakes  with  regard  to  the  signs  are  often  made  in 
clearing  an  equation   of  fractions.      In  this   equation   the  fraction 

— x  is  regarded  as  a  single  term  with  the  minus  sign  hefore  it:  it 

2 
is  equivalent  to  —  \{x  —  1).     When  multiplied  by  10,  it  is  well  to  put 
the  result  first  in  the  form  — 5(x  —  1),  and  afterwards  in  the  form 
_.V  _|_  5,  in  order  to  secure  attention  to  the  signs. 

2.  Solve  4  -  — —  =*--!.  Ans.  33. 

8  22        2 

3.  "      |(5a;  +  3)  -  £(16  -  5a;)  =  37  -  4a?.  6. 

4        M      Gx  +  15  _  Sx  -  10  =  4.v;  -  7  3 

11  7  5 

Note.  —  In  certain  cases  it  is  more  advantageous  not  to  clear  the 
equation  of  fractions  at  once  by  multiplying  it  throughout  by  the  least 
tommon  multiple  of  the  denominators  (Art.  55),  but  to  clear  it  of 
fractions  partially,  and  then  to  effect  some  reductions,  before  we 
semove  the  remaining  fractions. 


76  EXAMPLES. 

K     o  ,       x  —  4    ,    2s  -  3        5x  —  32        a;  -f-  9 

5.    Solve H —  = ^-— . 

3  35  9  28 

First  multiply  by  9,  and  we  have 

&  _  i2  +  18*  ~  27  =  5,,  _  3 2  _  9s  +  81 
35  28       ' 

18s  -  27   ,   9s  -f-  81       Q         on 
transposing, — 1 -1- —  =  2s  —  20. 

Now  clear  of  fractions  by  multiplying  by  the  least  common 
multiple  of  the  denominators,  which  is  5  x  7  x  4,  or  140, 
and  we  get 

72s  -  108  +  45s  +  405  ==  280s  -  2800  ; 
transposing,  72s  -f-  45s  -  280s  =  -2800  +  108  -  405; 
collecting  terms,  —  lG3s  =  —3097; 

dividing  by  —163,  x  =  19. 

Solve  the  following  equations : 

-       X  —  2     .     X  -f-   10  K  a  a 

6.   — h  — - —  =  5.  Ans.  8. 


r-         X    -f~     1«/    q       .       S 

5  4' 


16. 


8.  tli  =  1  +  *±I>  17. 

8  18 

q     4(x  +  2)  __  5s 

9.  —  -  7  +  -.  13. 

10.  »±J0  +  *JL  =  6.  7. 

9  7 

11.  ®JzJ*  +  ?LzJ?  +  A  -  o.  4. 

7       T       3       T  21 

12     *  +  f)  _  *  +  1  =  ^  +  3  1 

6  9  4*  7* 

13.  3-^^  -  ±(X  -  4)  =  !(.-  6)  +  A.  6. 

16  12v  '        5V  y       48 

14.  f  -  ±<«  +  10)  -  (s  -  3)  -  ^L_Z  _  4|.  7. 


EQUATIONS    WHOSE    COEFFICIENTS   ARE    DECIMALS.     77 

59.  To  Solve  Equations  whose  Coefficients  are 
Decimals,  it  is  advisable  generally  to  express  all  the 
decimals  as  common  fractions,  to  insure  accuracy,  and  then 
proceed  as  before  ;  but  it  is  often  found  more  simple  to 
work  entirely  in  decimals. 

EXAM  PLES. 

1.  Solve  .6*x  +  -25  -  Jaj  =  1.8  -  .75a:  -  -J. 

Expressing  the  decimals  as  vulgar  fractions,  we  have 

2r     i      1    _    lr   _    1  8    _    3  v   _    1  . 

clearing  of  fractions, 

24a:  -f  9  -  4a:  =  68  -  27a;  -  12  ; 
transposing,  24x  —  Ax  +  27a?  =  68  —  12  —  9 ; 
.-.     47as  =  47  ; 
x  =    1. 

2.  Solve  ,375a;  -  1.875  =  ,12a;  +  1.185. 
Transposing,  ,375a:  —  ,12a;  =     1.185  +  1.875; 
collecting  terms,  (.375  —  ,12)a:  =    3.06; 

that  is  ,255a:  =    3.0G  ; 

dividing  by  .255,  x  =  12. 

3.  Solve  ,5a:  —  ,3a:  =  ,25a:  —  1.  Ans.  12. 

4.  "      ,2a;  -  ,16a;  =  .6  -  .3.  8. 

5.  "      2.25a:  —  .125  =  3a:  A-  3.75.  —  5£. 

60.  Literal  Equations.  —  A  Literal  Equation  is  one  in 
which  some  or  all  the  numbers  are  represented  by  letters. 

Thus 

ax2  +  bx  =  ex  +  4,     and     ax  +  b  =  ex*  —  d, 

are  literal  equations.  The  known  numbers  are  usually 
represented  by  the  first  letters  of  the  alphabet,  as  «,  b,  c, 
etc. 

*  .6  denotes  the  repeteud  .6GG0  etc.  =  §.    Similarly  .8  denotes  .888  etc.  =  $ . 


78  LITERAL   EQUATIONS  —  EXAMPLES. 

EXAMPLES. 

1.  Solve  ax  4-  b2  =  &aj  4-  a2. 

Transposing,  we  have 

ax  —  bx  =  a2  —  Z>2, 

that  is  (a  —  &)se  =  a2  —  62; 

dividing  by  a  —  &,  the  coefficient  of  #,  we  have 

x  =  (a2  -  Z>2)  -5-  (a  -  6)  =  a  +  6. 

2.  Solve   X  4-  -  =  c. 

a        o 

Multiplying  by  a&,  we  have 

bx  +    aaj  =  abc, 
that  is,  (ft  4-  b)x  =  «&c  ; 

dividing  by  a  4-  6,  the  coefficient  of  jc,  we  have 

a  +  b 

3.  Solve  (a  -  x)  (a  4-  x)  =  2a2  4-  2aa;  -  x\ 

Ans.  x  =  — 


4.   Solve  2#  4-  bx  —  a  =  3a  —  2c.  a  = 


a 

2* 

a  -  2c 


&  -  1 
5.    Solve  ax  —  bx  4-  Z>2  =  a2.  a  =  a  4-  b, 

G.   Solve  (a  4-  x)  (&  4-  a?)  =  a(&  4-  c)  4-  —  4-  a2. 

^,9.  X  =   — . 

6 

7.  Solve  a(#  —  a)  4-  &(*  —  b)  =  2ab.  x  =  a  4-  &• 

8.  Solve  2(x  -  a)  4-  3(a>  -  2a)  =  2a.  a  =  2a. 

9.  Solve  J(su  +  a  +  i)+  \{x  +  a  -  b)  =  b. 

Ans.  x  =  b  —  a. 
10.   Solve  (a  4-  &«)(&  4-  as)  =  ab(x2  —  1). 

2«?j 
a2  4-  &*' 


PROBLEMS   LEADING    TO   SIMPLE   EQUATIONS.  79 

61.  Problems    Leading    to    Simple    Equations.— 

The  preceding  principles  may  now  be  employed  to  solve 
various  problems. 

A  Problem  is  a  question  proposed  for  solution.  In  a 
problem  certain  quantities  are  given  or  known,  and  certain 
others  which  have  some  assigned  relations  to  these,  are 
required. 

A  Theorem  is  a  truth  requiring  proof. 

Axioms,  Problems,  and  Theorems,  are  called  Propositions. 

The  Solution  of  a  problem  by  Algebra  consists  of  two  dis- 
trict parts  :  (1)  The  Statement  of  the  problem,  and  (2)  the 
Solution  of  the  equation  of  the  problem. 

The  Statement  of  the  problem  is  the  process  of  expressing 
the  conditions  of  the  problem  in  Algebraic  language  by  an 
equation.  The  statement  of  the  problem  is  often  more 
difficult  to  beginners  than  the  solution  of  the  equation.  Xo 
rule  can  be  given  for  the  statement  of  every  particular 
problem.  Much  must  depend  on  the  skill  of  the  student, 
and  practice  will  give  him  readiness  in  this  process.  The 
following  is  the  general  plan  of  finding  the  equation  : 

1.  Study  the  problem,  to  ascertain  what  quantities  in  it  are 
known  and  what  are  unknown,  and  to  understand  it  fully,  so 
as  to  be  able  to  prove  the  correctness  or  incorrectness  of  any 
proposed  answer. 

2.  Represent  the  unknown   quantity  by  one   of  the  final 

letters  of  the  alphabet,  say  x,  and  express  in  Algebraic 
language  the  relations  which  hold  between  the  known  and  un- 
known quantities;  an  equation  wiU  thus  be  obtained  which  can 
be  solved  by  the  methods  already  given,  and  from  which  the 
value  of  the  unknown  quantity  may  be  found. 

Xote  1.  —  Problems  may  often  involve  several  unknown  quantities, 
but  in  the  present  chapter  we  shall  consider  only  problems  in  which 
there  is  one  unknown  quantity,  or  in  which,  if  there  are  several,  they 
are  so  related  to  one  another  that  they  can  all  be  expressed  in  terms 
of  some  one  of  them. 


80  EXAMPLES. 


EXAMPLES. 

1.  What  number  is  that  whose  double  exceeds  its  half  by 
27? 

Let  x  represent  the  number  ; 
then  2x  represents  the  double  of  the  number, 

x 
aud  -  represents  the  half  of  the  number. 

Since   from   the   conditions   of    the   problem   the   double 
exceeds  the  half  by  27,  we  have  for  the  equation 

2x  -  -  =  27. 
2 

Clearing  of  fractions, 

4x  —  x  =  54, 
that  is,  3x  =  54. 

.-.  x  =  18. 
Hence  the  required  number  is  18. 

Verification,  2  x  18  -  —  =27. 
2 

2.  The  sum  of  two  numbers  is  28,  and  their  difference  is 


Let  x  =  the  smaller  number ; 
then  x  -|-  4  =  the  greater  number ; 

and  since,  from  the  conditions  of  the  problem,  the  sum  is  to 
be  equal  to  28,  we  have  for  the  equation 
x  +  x  +  4  =  28  ; 
that  is  2x  =  24. 

.*.     x  =  12, 
and  x  +  4  =  16, 

so  that  the  numbers  are  12  and  1G. 

Verification,  16  +  12  =  28,  and  16  —  12  =  4. 
The  beginner  is  advised  to  test  each  solution  by  prating 
that  it  satisfies  the  data  of  the  question. 

3.    A  has  $80  and  B  has  $15.     How  much  must  A  give  to 
1>  in  order  that  he  may  have  just  four  times  as  much  as  B? 


EXAMPLES.  81 

Let  x  =  the  number  of  dollars  that  A  gives  to  B  ; 

then  80  —  x  =  the  number  of  dollars  that  A  has  left, 
and   15  -f  x  =  the  number  of  dollars  that  B  will  have  after 
receiving  x  dollars  from  A. 

But  A  has  now  four  times  as  much  as  B  ;  hence  we  have 
the  equation 

80  -  x  =  4(15  +  x), 
that  is,  80  —  x  ==  60  +  Ax, 

transposing  and  uniting,  —ox  =  —20, 
dividing  by  —5,  x  =  4. 

Hence  A  must  give  84  to  B. 

4.  A  father  is  six  times  as  old  as  his  son,  and  in  four 
years  he  will  be  four  times  as  old.     How  old  is  each? 

Let         x  =  the  son's  age  in  j-ears, 
then         6x  =  the  father's  age  in  }Tears. 
Also  x  -f-  4  =  the  son's  age  in  years,  after  four  years, 
and  6o5  4-  4  =  the  father's  age  in  years,  after  four  years. 
Hence,  from  the  conditions,  we  have  the  equation 

6x  +  4  =  40  -f  4), 
that  is,  6a?  +  4  ==  4a;  +  16  ; 

.•.    x  =  6,  the  son's  age, 
and  6#  =  36,  the  father's  age. 

5.  Divide  60  into  two  parts,  so  that  3  times  the  greater 
may  exceed  100  by  as  much  as  8  times  the  less  falls  short  of 
200. 

Let  x  =  the  greater  part, 

then  60  —  x  =  the  less. 

Also  Sx  —  100  =  the  excess  of  3  times  the  greater  over 
100,  and  200  —  8(60  —  x)  =  the  number  that  8  times  the 
less  falls  short  of  200.  Hence,  from  the  conditions,  we 
have  the  equation 

3a;  -  100  =  200  -  8(60  -  x), 
that  is,  3a;  -,100  =  200  -  480  -f-  8a;, 

hence,  — oa;  =  —180. 

.-.    x  =  36,  the  greater  part. 
60  —  X  =  24,  the  less. 


82  EXAMPLES. 

6.  A  line  is  2  feet  4  inches  long ;  it  is  required  to  divide 
it  into  two  parts,  such  that  one  part  may  be  three-fourths  of 
the  other  part. 

Let     x  =  the  number  of  inches  in  the  larger  part, 
then      fas  =  the  number  of  inches  in  the  other  part. 
Hence,  from  the  conditions,  we  have  the  equation 

x  +  %x  =  28, 
that  is,  ±x  +  ox  =  112. 

.-.     x  =  16. 
Thus  one  part  is  16  inches  long,  and  the  other  part  12  inches 
long. 

7.  Divide  $47  between  A,  B,  C,  so  that  A  may  have  $10 
more  than  B,  and  B  $8  more  than  C. 

Note  2.  —  Here  there  are  really  three  unknown  quantities,  but  it 
is  only  necessary  to  represent  the  number  of  dollars  the  last  has  by  a 
symbol. 

Let  x  =  the  number  of  dollars  that  C  has, 

then  x  -f  8  =  the  number  of  dollars  that  B  has, 

and    x  +  8  -f-  10  =  the  number  of  dollars  that  A  has. 
Hence  we  have  the  equation 

x  +  (x  +  8)  +  (x  -f-  8  -f  10)  =  47, 
.-.      3a;  =  21, 

x  ==  7; 

so  that  C  has  $7,  B  $15,  A  $25. 

8.  A  person  spent  £28.  4s.  in  buying  geese  and  ducks  ;  if 
each  goose  cost  7s.,  and  each  duck  cost  3.9. ,  and  if  the  total 
number  of  birds  bought  was  108,  how  many  of  each  did  he 
buy? 

Note  3. — In  questions  of  this  kind  it  is  of  essential  importance 
to  have  all  concrete  quantities  of  the  same  kind  expressed  in  the  same 
denomination;  in  the  present  instance  it  will  be  convenient  to  express 
the  money  in  shillings.  In  Ex.  G  it  was  convenient  to  express  the 
length  in  inches. 

Let  x  =  the  number  of  geese, 

then        108  —  x  =  the  number  of  ducks. 

Also  7a?  =  the  number  of  shillings  the  goose  cost, 

and    3(108  —  x)  =  the  number  of  shillings  the  ducks  cost 


EXAMPLES.  83 

But  from  the  conditions  of  the  question,  the  whole  cost  of 
the  geese  and  ducks  is  £28.  46-.,  i.e.,  0G4  shillings,  lleuce 
we  have  the  equation 

7x  +  3(108  -  x)  =  564, 
that  is,    7x  -f  324  —  ox  =  564, 

.-.     x  =  60,  the  number  of  geese, 
and  108  —  x  =  48,  the  number  of  ducks. 

9.  A  can  do  a  piece  of  work  in  12  hours,  which  B  can  do 
in  4  hours.  A  begins  the  work,  but  after  a  time  B  takes  his 
place,  and  the  whole  work  is  finished  in  6  hours  from  the 
beginning.     How  long  did  A  work  ? 

Let  x  =  the  number  of  hours  that  A  worked, 

then      6  —  x  =  the  number  of  hours  that  B  worked. 

Also  j3^  =  the  part  A  does  in  1  hour,  since  he  can  do 

the  whole  work  in  12  hours. 

x 
Therefore  —  =  the  part  done  by  A  altogether. 

Also  J  =  the  part  B  does  in  1  hour,  since  he  can  do 

the  whole  work  in  4  hours. 
Therefore  J  (6  —  x)  =  the  part  done  by  B  altogether. 

But  A  and  B  together  do  the  ivhole  work  ;  hence  the  sum 
of  the  parts  of  the  work  that  they  do  separately  must  equal 
unity;  and  we  have  for  the  equation 

—  +  -(6  -  x)  =  1. 
12       4V  J 

Multiplying  by  12,  we  have 

x  +  3(6  —  x)  =     12, 

...     _2aj  =  -6. 

x  =      3. 

Hence  A  worked  for  3  hours. 

Note  4.  —  It  should  be  remembered  that  x  must  always  represent 
a  number;  what  is  called  the  unknown  quantity  is  really  an  unknown 
number.  In  the  above  examples  the  unknown  quantity  x  represents 
a  number  of  dollars,  years,  inches,  etc.  For  instance,  in  Ex.  6,  we  let 
x  denote  the  number  of  inches  in  the  longer  part;  beginners  often  say, 


84  EXAMPLES 

"  let  x  =  the  longer  part,"  or,  "  let  x  —  a  part,"  which  is  not  definite, 
because  a  part  may  be  expressed  in  various  ways,  in  feet,  or  inches,  or 
yards.  Again,  in  Ex.  7,  we  let  x  =  the  number  of  dollars  that  C  has; 
beginners  often  say,  "let  x  =  C's  money,"  which  is  not  definite, 
because  C's  money  may  be  expressed  in  various  ways,  in  dollars,  or  in 
pounds,  or  as  a  fraction  of  the  whole  sum.  The  student  must  be 
careful  to  avoid  beginning  a  solution  with  a  vague  and  inexact 
statement. 

It  may  seem  to  the  student  that  some  of  the  problems  which  are 
given  for  exercise  can  be  readily  solved  by  Arithmetic,  and  he  may 
therefore  be  inclined  to  undervalue  the  power  of  Algebra  and  consider 
it  unnecessary.  We  may  remark,  however,  that  by  Algebra  the  student 
is  enabled  to  solve  all  the  problems  given  here,  without  any  uncer- 
tainty; and  also,  he  will  find  as  he  proceeds,  that  he  can  solve 
problems  by  Algebra,  which  would  be  extremely  difficult  or  entirely 
impracticable,  by  Arithmetic  alone. 

10.  The  difference  between  two  numbers  is  8  ;  if  2  be 
added  to  the  greater  the  result  will  be  three  times  the 
smaller:  find  the  numbers.  Ans.  13,  5. 

11.  A  man  walks  10  miles,  then  travels  a  certain  distance 
by  train,  and  then  twice  as  far  by  coach.  If  the  whole 
journey  is  70  miles,  how  far  does  he  travel  by  train? 

Ans.  20  miles. 

12.  What  two  numbers  are  those  whose  sum  is  58,  and 
difference  28?  Ans.  15,  43. 

13.  If  288  be  added  to  a  certain  number,  the  result  will 
be  equal  to  three  times  the  excess  of  the  number  over  12  : 
find  the  number.  Ans.  162. 

14.  Find  three  cousecutive  numbers  whose  sum  shall 
equal  84.  Ans.  27,  28,  29. 

15.  Find  two  numbers  differing  by  10,  whose  sum  is  equal 
to  twice  their  difference.  Ans.  15,  5. 

10.  Find  a  number  such  that  if  5,  15,  and  35  are  added 
to  it,  the  product  of  the  first  and  third  results  may  be  equal 
to  the  square  of  the  second.  Ans.  5. 

17.  A  is  twice  as  old  as  B,  and  seven  years  ago  their 
united  ages  amounted  to  as  many  years  as  now  represent  the 
age  of  A :  find  the  ages  of  A  and  B.  Ans,  28,  14. 


EXAMPLES.  85 


EXAMPLLS. 

Solve  the  following  equations  : 

1.  3.6-  -\r  -.5  =  x  -j-  25.  Ans.  5. 

2.  2a;  +  3  =  16  -  (2x  -  3).  4. 

3.  7(25  -  x)  -  2x  =  2(3a;  -  25).  15. 

4.  5x  —  17  +  Sx  —  5  =  6a;  —  7  —  8a;  +  115.  13. 

5.  5(aJ  -f-  2)  =  3(>  -f  3)  +  1.  0. 
G.  2(x  -  3)  =  5(as  +  1)  +  2x  -  1.  -2. 

7.  2(a?  -  1)  -  3(a;  -  2)  +  4(a;  -  3)  +  2  =  0.  2. 

8.  bx  +  6(a;  +  1)  -  7(a;  +  2)  -  8(a;  +  3)  =  0.       -8. 

9.  (.v  -f  l)(2a;  +  1)  =  (x  +  3)  (2a;  +  3)  -  14.  1. 

10.  O+l)2-(a;2-l)=a;(2a;+l)-2O-f2)(a;+l)+20. 

Ans.  2. 

11.  G(a;2-3a;+2)-2(a;2-l)=4(a;+l)(a;+2)--24.  1. 

12.  2x  -  o\3x  -  7  (4a;  -  9)  \  =  66.  3. 

13.  3(5  -  6a;)  -  5[x  -  5|1  -  3(a;  -  5)  J]  =  23.  4. 

14.  (x  +  l)2  +  2(x  +  3)2  =  3a;(a;  +  2)  +  35.  2. 

15.  84+0+4)(a;-3)(a;+5)  =  (a;+l)(a;+2)(^+3).  1. 
1G.  (x-\-l)(x  +  2)(x  +  6)  =  x3+()x2  +  4(7x-  1).  2. 

17.  5  _  £  _  i.  _20. 
5       4 

18.  ?-— 5  +  ^-=-2  =  3.  5. 

2  3 

19.  }(«  +  1)  -  f  (a;  -  1)  =  3.  -5. 

20.  J(2  -  x)  -  |(&B  +  21)  =  x  +  3.  -2Jf. 
21     t±l  +  i+J+i±i  +  8sB,a                    _9A. 

22.  *=J  -  2-ZLi  =  -3LziJ  _  rx  _  2).  2± 

2  3  2  V  ;  2 

23.  ^-±-1  -  2x  ~  i  +  H  =  0.  -16. 

2  5  4 

24.  8*  +  5  -  21  +  *  =  5*  -  15.  1. 


25.    2  -  T\(x  -  11)  =  f(a;  -  25)  +  34.  25. 


86 


EXAMPLES. 


26.    l(x  -  8)  +  ^4^  +  ^-=-^  =  7  -  2 


27,    *  -  (to  -  ^j~j  =  *(&*  -  S7)  -  f. 


5 

-4ns.  8. 


28. 

1  -  2x 
3 

29. 

x  4  1 
3 

30. 

x  +  3 

2 

31. 

3a?  -  1 

5 

32. 

2-x 
3 

33. 

5a;  —  3 

i)X 


G 


rr:  _i_  is  o 

'      42    —    w- 


4  4aj  =  12  4  2x  ~  1, 


3x  -  5 


3  12        "    * 

4  +       5       +       G       +^-U* 

7  3  2         ^  v  J 

34.  fa;  +  .25a;  -  .3x  =  x  —  3. 

35.  .5a  —  .2a  =  .Sx  —  1.5. 

3G.    1.5  =  ^--09*--18 
.2  .9 

37.  (a  +  b)x  4  (a  -  b)x  =  a2. 

38.  (a  4  b)x  4  (6  -  a)  a  =  62. 
30.   J  (a  4  a?)  4  Wa  4  x)  4  }(3a  +  »)  =  3a. 

40.  S  4  -  =  a2  +  62.  ab. 

0         a 

41.  (a2  4-  x)(b2  4  ®)  =  (ab  4  ^)2.  0. 

42.  a(x  +  a)  +  6(6  -  a)  =  2ab.  b  -  a. 

43.  ax(x  +  a)  4  bx(x  +  b)  =  (a  4  &)  (oj  4  a)  (a  4  &). 

ylws.  —  £  (a  +  &). 

44.  («-a)>+(*-6)«+(»-c)>=8(»-o)(*-6)(»-c). 

Ans.  |(a  4-  6  4-  c). 

45.  Twenty-three  times  a  certain  number  is  as  much  above 
14  as  1G  is  above  seven  times  the  number:  find  it.       Ans.  1. 


EXAMPLES.  87 

46.  Divide  105  into  two  parts,  one  of  which  diminished 
by  20  shall  be  equal  to  the  other  diminished  by  15. 

Ans.  50,  55. 

47.  The  sum  of  two  numbers  is  8,  and  one  of  them  with 
22  added  to  it  is  five  times  the  other :  find  the  numbers. 

Ans.  3,  5o 

48.  A  and  B  begin  to  play  each  with  860.  If  they  play 
till  A's  money  is  double  B's,  what  does  A  win?       Ans.  $20. 

49.  The  difference  between  the  squares  of  two  consecutive 
numbers  is  121  :  find  the  numbers.  Ans.  60,  61. 

50.  Divide  8380  between  A,  B,  and  C,  so  that  B  may 
have  $30  more  than  A,  and  C  may  have  $20  more  than  B. 

Ans.  A  6100,  B  8130,  C  $150. 

51.  A  father  is  four  times  as  old  as  his  son  ;  in  24  years 
he  will  only  be  twice  as  old  :  find  their  ages.       Ans.  48,  12. 

52.  A  is  25  years- older  than  B,  and  A's  age  is  as.  much 
above  20  as  B's  is  below  85  :  find  their  ages.      Ans.  65,  40. 

53.  The  sum  of  the  ages  of  A  and  B  is  30  years,  and  five 
years  hence  A  will  be  three  times  as  old  as  B :  find  their 
present  ages.  Ans.  25,  5. 

54.  The  length  of  a  room  exceeds  its  breadth  by  3  feet ; 
if  the  length  had  been  increased  by  3  feet,  and  the  breadth 
diminished  by  2  feet,  the  area  would  not  have  been  altered  : 
find  the  dimensions.  Ans.  15  ft.,  12  ft. 

55.  There  is  a  certain  fish,  the  head  of  which  is  9  inches 
long ;  the  tail  is  as  long  as  the  head  and  half  the  body  ;  and 
the  body  is  as  long  as  the  head  and  tail  together  :  what  is  the 
length  of  the  fish  ?  Ans.  6  ft. 

56.  The  sum  of  $76  was  raised  by  A,  B,  and  C  together ; 
B  contributed  as  much  as  A  and  $10  more,  aud  C  as  much 
as  A  and  B  together :  how  much  did  each  contribute? 

Ans.  $14,  $24,  $38. 

57.  After  34  gallons  had  been  drawn  out  of  one  of  two 
equal  casks,  and  80  gallons  out  of  the  other,  there  remained 
just  three  times  as  much  in  one  cask  as  in  the  other :  what 
did  each  cask  contain  when  full?  Ans.  103. 


88  EXAMPLES. 

58.  Divide  the  number  20  into  two  parts  such  that  the 
sum  of  three  times  one  part,  and  five  times  the  other  part, 
may  be  84.  Ans.  8,  12. 

59.  A  person  meeting  a  company  of  beggars  gave  4  cents 
to  each,  and  had  1(3  cents  left;  he  found  that  he  should 
have  required  12  cents  more  to  enable  him  to  give  the 
beggars  6  cents  each :    how  many  beggars  were  there  ? 

Ans.  14. 

GO.  Divide  100  into  two  parts  such  that  if  a  third  of  one 
part  be  subtracted  from  a  fourth  of  the  other,  the  remainder 
may  be  11.  Ans.  24,  76. 

Gl.  Divide  60  into  two  parts  such  that  the  difference 
between  the  greater  and  64  may  be  equal  to  twice  the 
difference  between  the  less  and  38.  Ans.  36,  24. 

62.  Find  a  number  such  that  the  sum  of  its  fifth  and  its 
seventh  shall  exceed  the  sum  of  its  eighth  and  its  twelfth  by 
113.  Ans.  840. 

63.  An  army  in  defeat  loses  one-sixth  of  its  number  in 
killed  and  wounded,  and  4000  prisoners  ;  it  is  re-enforced 
by  3000  men,  but  retreats,  losing  one-fourth  of  its  number 
in  doing  so  ;  there  remain  18000  men  :  what  was  the  original 
force?  Ans.  30000. 

64.  One-half  of  a  certain  number  of  persons  received  18 
cents  each,  one-third  received  24  cents  each,  and  the  rest 
received  30  cents  each ;  the  whole  sum  distributed  was 
$5.28  :  how  many  persons  were  there?  Ans.  24. 

65.  A  father  has  six  sons,  each  of  whom  is  four  years 
older  than  his  next  younger  brother ;  and  the  eldest  is  three 
times  as  old  as  the  youngest:  find  their  respective  ages. 

Ans.  10,  14,  18,  22,  26,  30. 

66.  A  man  left  his  property  to  be  divided  between  his 
three  children  in  such  a  way  that  the  share  of  the  eldest  was 
to  be  twice  that  of  the  second,  and  the  share  of  the  second 
twice  that  of  the  youngest ;  it  was  found  that  the  eldest 
received  $3000  more  than  the  youngest:  how  much  did  each 
receive?  Ana.  $1000,  $2000,  $1000. 


EXAMPLES. 

67.  A  sum  of  money  is  divided  among  throe  persons  ;  the 
first  receives  810  more  than  a  third  of  the  whole  sum ; 
the  second  receives  815  more  than  a  half  of  what  remains ; 
and  the  third  receives  what  is  over,  which  is  870  :  find  the 
original  sum.  Ans.  8270. 

68.  In  a  cellar  one-fifth  of  the  wine  is  port  and  one-third 
claret;  besides  this  it  contains  15  dozen  of  sherry  and  30 
bottles  of  spirits  :  how  much  port  and  claret  does  it  contain  ? 

Ans.  90  port,  150  claret. 

69.  Two-fifths  of  A's  money  is  equal  to  B's,  and  seven- 
ninths  of  B's  is  equal  to  C's  :  in  all  they  have  8770,  what 
have  they  each?  Ans.  A  8450,  B  8180,  C  8140. 

70.  A,  B,  and  C  have  $1285  between  them  ;  A's  share  is 
greater  than  five-sixths  of  B's  by  82-3,  and  C's  is  four- 
fifteenths  of  B's  :   find  the  share  of  each. 

Ans.  A  8525,  B  8600,  C  $160. 

71.  A  sum  of  money  is  to  be  distributed  among  three 
persons,  A,  B,  and  C  ;  the  shares  of  A  and  B  together 
amount  to  S240  ;  those  of  A  and  C  to  8320  :  and  those  of  B 
and  C  to  $368  :  find  the  share  of  each  person. 

Ans..  $96,  8144,  8224. 

72.  Two  persons  A  and  B  are  travelling  together;  A  has 
81 00,  and  B  has  $48  :  they  are  met  by  robbers  who  take 
twice  as  much  from  A  as  from  B,  and  leave  to  A  three  times 
as  much  as  to  B :  how  much  was  taken  from  each? 

Ans.  888.  844. 

73.  In  a  mixture  of  wine  and  water  the  wine  composed 
25  gallons  more  than  half  of  the  mixture,  and  the  water  5 
gallons  less  than  a  third  of  the  mixture  :  how  many  gallons 
were  there  of  each?  Ans.  85,  35. 

74.  A  general,  after  having  lost  a  battle,  found  that  he 
had  left  fit  for  action  3600  men  more  than  half  of  his  army  : 
600  men  more  than  one-eighth  of  his  army  were  wounded  ; 
and  the  remainder,  forming  one-fifth  of  the  army,  were  slain, 
taken  prisoners,  or  missing  :  what  was  the  number  of  the 
army?  Ans.  24000. 


00      WHEN  ALL    TERMS  HAVE   ONE   COMMON  FACTOR. 


CHAPTER    VII. 

FACTORING  — GREATEST    COMMON    DIVISOR- 
LEAST     COMMON     MULTIPLE. 

62.  Definitions.  —  Factoring  is  the  process  of  resolving 
a  quantity  into  its  factors. 

The  Factors  of  a  quantity  are  those  quantities  which 
multiplied  together  produce  it.  A  factor  of  a  quantity  is 
therefore  a  divisor  of  the  quantity,  i.e.,  it  will  divide  the 
quantity  without  a  remainder.  Thus,  a  is  a  factor  or  divisor 
of  abc,  and  b  is  a  factor  or  divisor  of  ab  —  b2. 


Note.  —In  Division  (Chap.  Y.)  we  had  given  the  product  of  two 
factors  and  one  of  the  factors,  and  we  showed  how  to  find  the 
other  factor.  In  the  present  chapter  we  shall  consider  cases  in  which 
the  factors  of  an  expression  can  he  found  when  none  of  the  factors 
are  given. 

A  Prime  Quantity  is  one  which  has  no  integral  factor 
except  itself  and  unity.  Thus,  o,  b,  and  a  +  c  are  prime 
quantities  ;  while  ab,  and  ac  +  be  are  not  prime. 

Quantities  are  said  to  be  prime  to  each  other  or  relatively 
prime,  when  unity  is  the  only  integral  factor  common  to 
both.     Thus,  ab  and  cd  are  prime  to  each  other. 

A  Composite  Quantity  is  one  which  is  the  product  of  two 
or  more  integral  factors,  neither  of  which  is  unity  or  the 
quantity  itself.  Thus,  ax  +  x*  is  a  composite  quantity, 
the  factors  of  which  are  as  and  a  -f-  X. 

63.  When  All  the  Terms  have  one  Common 
Factor.  —  When  each  term  of  :i  polynomial  is  divisible 
by  a  common  factor,  the  polynomial  may  be  simplified  by 
the  following 


EXAMPLES.  91 

KULE. 

Divide  each  term  of  the  polynomial  separately  by  the  com- 
mon factor,  and  enclose  the  quotient  within  pareritheses,  the 
common  factor  being  placed  outside  as  a  coefficient;  then  the 
divisor  will  be  one  factor  and  the  quotient  the  other. 

EXAMPLES. 

1 .  Factor  the  expression  3a2  —  Gab. 

Here  we  see  that  the  terms  have  a  common  factor,  3a ; 
therefore,  dividing*  the  polynomial  by  3a,  we  obtain  for  the 
quotient  a  —  26.  Hence  the  two  factors  are  3a  and  a  —  26. 
.-.     3a2  -  Gab  =  3a(a  -  26). 

Similarly 

2.  5a26a4  -  IhaXM  -  20ab*x*  =  5abx*(ax  -  36  -  462a). 
Factor  the  following  expressions  : 


3. 

x2  —  ax. 

Ans.  x(x  —  a). 

4. 

Xs   —   X2. 

x\x  -  1). 

5. 

a2  —  ab2. 

a(a  -  62). 

6. 

8x  -  2x2. 

2a(4  -  x). 

7. 

5  ax  —  5a3a2. 

5ax(l  —  a2a). 

8. 

x3  —  x2y. 

x2(x  -  y). 

9. 

5x  —  25x2y. 

5a  (1  —  bxy). 

10. 

Wx2  +  Ux2y. 

16a2(l  +  4y). 

11. 

54  -  81a. 

27(2  -  3a). 

12. 

3a3  -  Qx2  +  9a. 

Sx(x2  -  2x  +  3). 

© 

6a26a3  +  2a62a4  +  4a6a5. 

2a6a3(3a  -f  bx  +  2a2). 

14. 

72a2?/  -  84a?/2  +  60^. 

12a?/(6a  —  7?/  +  5a?/). 

64.  Expressions  containing  Four  Terms.  —  When  a 
pol}Tnomial  contains  four  terms  which  can  be  arranged  in 
pairs  that  have  a  common  binomial  factor,  the  polynomial 
may  be  simplified  by  the  following 

Rule. 

Divide  the  polynomial  by  the  common  binomial  factor;  then 
the  divisor  will  be  one  factor  and  the  quotient  the  other. 


92  EXAMPLES. 

EXAMPLES. 

1 .  Resolve  into  factors  a;2  —  ax  4-  bx  —  ab. 

Here  we  see  that  the  first  two  terms  contain  a  factor  a;,  and 
the  last  two  terms  a  factor  b  ;  therefore  we  factor  the  first 
two  and  last  two  terms  by  Art.  63,  and  obtain  x(x  —  a) 
and  b(x  —  a).  We  now  see  that  the  two  pairs  have  the 
common  binomial  factor  x  —  a.  Dividing  by  x  —  a  we 
obtain  the  quotient  x  -f-  o  f°r  the  other  factor. 
The  work  therefore  will  stand  as  follows : 

x2  —  ax  +  bx  —  ab  =  x(x  —  a)  4  b(x  —  a) 
—  (x  —  a)(x  4-  b). 

2.  Resolve  into  factors  6a;2  —  dax  -f-  46a;  —  Gab. 

Gx2  —  9ax  +  46a;  —  Gab  =  3x(2x  —  3a)  +  2b{2x  —  3a) 
=  (2x  -  3a)  (3a;  +  26). 

3.  Resolve  into  factors  12a2  —  4a6  —  3aa;2  +  bx2. 
12a2  -  4a&  -  3aa;2  +  6a;2  =  4a (3a  -  6)  -  a;2(3a  -  6) 

=  (3a  -  6)  (4a  -  x2). 

Note.  — It  is  not  necessary  always  to  factor  in  the  same  way.  In 
the  first  line  of  work  it  is  usually  sufficient  to  see  that  each  pair 
contains  some  common  factor;  and  any  suitably  chosen  pairs  will 
bring  out  the  same  result.  Thus,  in  the  last  example,  we  may  have 
a  different  arrangement,  and  enclose  the  first  and  third  terms  in  one 
pair,  and  the  second  and  fourth  in  another  as  follows: 

12a'2  -  Aab  -  Sax*  +  bx2  =  ]2a2  -  Sax2  -  (4a6  -  bx2) 
=  3a (4a  —  a;2)  -  b (4a  -  x2) 
=  (4a  -  x2)  (3a  -  b), 
which  is  the  same  result  as  before. 

Resolve  into  factors 

4£.{dl  4-  a6';4-  ac  +  6c.  -4ns.  (a  +  6)  (a  +  c) 

5.    a2  —  ac  -f  ab  —  be.  (a  —  c)(a  +  6) 

G.    a2c2  +  acd  +  abc  4-  bd.  (ac  4  d)(ac  4-  6) 

a2  +  3a  4-  ac  4-  3c.  (a  +  3)(aT  c) 

2aa;  4-  ay  +  26a;  4-  by,  (2x  4-  y)  (a  4-  6) 

3ai»  —  6a;  —  Bay  4  6y.  (3a  —  6)  (x  —  ?/) 

aa;2  +  6a;2  4  2a  4  26.  (a  4-  6)  (a;2  4  2) 

a;2  -  3a;  -  xy  4  8y,  (a;  -  3)  (as  -  y) 


TO  FACTOR  A  TRINOMIAL  OF  THE  FORM  flJa  +  ax  +  6.    03 

65.  To  Factor  a  Trinomial  of  the  Form  x2+ax+b. — 
Let  x2  -{-  ax  +  b  be  any  trinomial  in  which  the  coefficient 
of  x2  is  +  1 ,  and  the  signs  of  a  and  b  either  plus  or  minus. 

Before  proceeding  to  explain  this  case  of  resolution  into 
factors,  the  student  is  advised  to  refer  to  Art.  40,  and 
examine  the  relation  that  exists  be  /ween  two  binomial 
factors  and  their  product.  Attention  was  there  called  to 
the  way  in  which,  in  forming  the  product  of  two  binomials, 
the  coefficients  of  the  different  terms  combined  so  as  to  give 
a  trinomial  result. 

Therefore,  in  the  converse  problem,  namely,  the  resolution 
of  a  trinomial  expression  into  its  component  binomial  factors, 
we  see,  by  reversing  the  results  of  Art.  40,  that  any  trino- 
mial may  be  resolved  into  two  binomial  factors,  when  the 
first  term  is  a  square,  and  the  coefficient  of  the  second  term 
is  the  sum  of  two  quantities  whose  product  is  the  third  term. 
Hence  the  following 

Rule 

The  first  term  of  each  factor  is  x,  and  the  second  terms 
are  two  numbers  whose  Algebraic  sum  is  the  coefficient  of  the 
second  term,  and  whose  product  is  the  third  term. 

The  application  of  this  rule  will  be  easily  understood  from 
the  following  ^ 

EXAMPLES. 

1.  Resolve  into  factors  x2  -f-  11a;  +  24. 

Here  the  first  term  of  each  binomial  factor  is  x,  and  the 
second  terms  of  the  two  binomial  factors  must  be  two 
numbers  whose  sum  is  11  and  whose  product  is  24.  It  is 
clear  therefore  that  they  must  be  -f-S  and  +3,  since  these 
are  the  only  two  numbers  whose  sum  is  11  and  whose  product 
is  24. 

.-.     x2  +  11a;  +  24  =  (x  +  S)(x  +  3). 

2.  Resolve  into  factors  x2  —  7x  +  12. 

The  first  term  of  each  factor  is  .t,  and  the  second  terms 
of  the  factors  must  be  such  that  their  sum  is  —7,  and  their 


04  EXAMPLES. 

product  is  +12.     Hence  they  must  both  be  negative,  and  it 
is  easy  to  see  that  they  must  be  —4  and  —3. 

.-.     x2  -  7x  +  12  =  (x  -  4)  (x  -  3). 

3.  Resolve  into  factors  x2  +  5x  —  24. 

The  first  term  of  each  factor  is  x,  and  the  second  terms 
of  the  factors  must  be  such  that  their  Algebraic  sum  is  -f- 5, 
and  their  product  is  —24.  Hence  they  must  have  opposite 
signs,  and  the  greater  of  them  must  be  positive  in  order  to 
give  the  positive  sign  to  their  sum.  It  is  easy  to  see  there- 
fore that  they  must  be  +8  and  —3. 

.-.     x2  +  hx  -  24  =  (x  +  8)0  -  3). 

4.  Resolve  into  factors  x2  —  x  —  56. 

The  first  term  of  each  factor  is  x,  and  the  second  terms 
of  the  factors  must  be  such  that  their  Algebraic  sum  is  —1, 
and  their  product  is  —56.  Hence  they  must  have  opposite 
signs,  and  the  greater  of  them  must  be  negative  in  order  to 
give  its  sign  to  their  sum.  The  required  terms  are  therefore 
-8  and  +7. 

.-.     x2  -  x  —  56  =  (a;  -  8)0  +  7). 

Note.  — In  examples  of  this  kind  the  student  should  always  verify 
his  results,  by  forming  the  product,  mentally,  of  the  factors  he  has 
chosen,  as  in  Art.  40. 


Resolve  into  factors 


5. 

a2  +  3a  +  2. 

6. 

x2  -  llx  +  30. 

7. 

a2  -  7a  +  12. 

8. 

x2  -  15.v  +  56. 

9. 

x2  —  19a;  +  90. 

10. 

x2  +  x  -  2. 

11. 

x2  +  x  -  6. 

12. 

x2  —  2x  -  3. 

13. 

x2  +  2x  -  3. 

14. 

x2  +  x  -  56. 

15. 

x-  +  \)x  —  40. 

Ans.  (a  +  l)(a  +  2) 
(x  -  6)0  -  5) 
(a  -4)(a-8) 

(x-8)(x-  7) 

O  -  9)  (x  -  10) 

0  +  2)0-  1) 

0  +  3)0-  -) 

O  -  3)0  +  1) 
0  +  »)(•-  1) 

O  +  8)0  -  7) 

0  +  8)0-  -r>) 


TO  FACTOR  A  TRINOMIAL  OF  THE  FORM  UX'  +  bx+C.      95 

16.  x2  —  4a?  —  12.  Ans.  (x  -  G)(<c  +  2). 

17.  x2  -  x  -  20.  {x  -  5)  (a;  +  4). 
S  a2  -  4a  -  21.  (a  -  7)  (a  +  3). 
1&  a2  +  a  -  20.  (a  +  5)  (a  -  4). 

52    a2  -  4a  -  117.  (a  -  13)  (a  +  9). 

@.    a2  -f  9a;  -  3G.  (a;  +  12)  (a;  -  3). 

Note.  —  If  the  term  containing  x-  is  negative,  enclose  the  whole 
expression  in  a  parenthesis  with  the  minus  sign  prefixed.  Then  factor 
the  expression  within  the  parenthesis  as  in  the  preceding  examples, 
and  change  the  signs  of  all  the  terms  of  one  of  the  factors.     Thus 

22.  Factor  90  +  9a;  -  x2. 
90+9a;-aJ2=-(a;2-9a;-90)  =  -(a;-15)(aj+6)  =  (15-a;)(ar+6). 

23.  Factor  240  +  x  —  x2.  Ans.  (16  —  a?)  (a;  +  15). 

24.  Factor  85  -f-  12a;  -  a;2.  (17  -  x)  (x  +  5). 
/||.  Factor  110  -  x  -  x2.  (10  -  x)(x  +  11). 
^26)   Factor  152  +  11a;  -  a;2.                   (19  -  x)\x  +  8). 

66.  To  Factor  a  Trinomial  when  the  Coefficient 
of    the    Highest    Power   is   not   Unity.  —  Let  it  be 

required  to  factor  3a;2  4-  14a;  +  8. 

The  first  term  3a;2  is  the  product  of  3a;  and  x. 

The  third  term  8  is  the  product  of  2  and  4  or  1  and  8. 

The  middle  term  14a;  is  the  result  of  adding  together  the 
two  products  3a;  and  2  and  x  and  4,  or  3a;  and  4  and  x  and  2, 
or  3a;  and  1  and  x  and  8,  or  3a;  and  8  and  x  and  1. 

Taking  the  first  products  we  have  3a;  X  2  +  x  x  4  =  10a; ; 
this  combination  therefore  fails  to  give  the  correct  middle 
term. 

Next  try  the  second  products,  and  get  3a;  x  4-f-  x  X  2  =  14a;, 
which  is  the  correct  value  of  the  middle  term. 

.-.     3a;2  +  14a;  +  8  =  (3a;  +  2)(x  +  4). 

The  beginner  will  frequently  find  that  it  is  not  easy  to 
select  the  proper  factors  at  the  first  trial.  Practice  alone 
will  enable  him  to  detect  at  a  glance  whether  any  two  factors 
are  the  proper  ones. 


96  EXAMPLES. 

EXAMPLES. 

1.  Resolve  into  factors  14a;2  +  29a;  —  15. 

Y\rrite  down  (7a;  5)  (2a;  3)  foi  a  first  trial,  noticing  that 
3  and  5  must  have  opposite  signs.  These  factors  give 
14a?2  and  —15  for  the  first  and  third  terms.  But  since 
7x3—  2x5  =  11,  this  combination  fails  to  give  the 
correct  coefficient  of  the  middle  term. 

Next  try  {lx  3)  (2x  5). 

Since  7x5  —  3x2  =  29,  these  factors  will  be  correct 
if  we  insert  the  signs  so  that  the  positive  will  predominate. 
.-.     14a;2  -f  2dx  -  15  =  (7a;  -  3)  (2a;  +  5). 

(Verify  by  mental  multiplication). 

It  is  not  usually  necessary  to  put  down  all  these  steps. 
After  a  little  practice  the  student  will  be  able  to  examine  the 
different  cases  rapidly,  and  to  reject  the  unsuitable  combina- 
tions at  once. 

In  the  factoring  of  such  expressions  as  these  the  following 
hints  are  very  useful : 

(1)  If  the  third  term  of  the  trinomial  is  positive,  then  the 
second  terms  of  its  factors  have  both  the  same  sign  as  the 
middle  term  of  the  trinomial. 

(2)  If  the  third  term  of  the  trinomial  is  negative,  then 
the  second  terms  of  its  factors  have  opposite  signs. 

2.  Resolve  into  factors  ox2  +  17a;  +  G (1) 

5a;2  —  llx  +  6 (2) 

5a;2  +  13a;  -  6 (3) 

5a;2  -  13a;  -  6 (4) 

In  (1)  we  notice  that  the  factors  which  give  G  are  both 
positive. 

In  (2)  we  notice  that  the  factors  which  give  G  are  both 
negative. 

In  (3)  we  notice  that  the  factors  which  give  6  have 
opposite  si;j;ns. 

In  (4)  wo  notice  that  the  factors  which  give  G  have 
opposite  signs. 


EXAMPLES.  07 

Therefore  for  (1)  we  write  (5a:  -f-    )(x  +  ). 

(2)  we  write  (5a:  —    )  (a;  —   ). 

(3)  we  write  (53      2)  (a;      3),  noticing  that 

2  and  3  have  opposite  signs. 

(4)  we  write  (5a      2)  (a:      3) ,  noticing  that 

2  and  3  have  opposite  signs. 
Since  5x3  +  1   X  2  =  17,  we  see  that 

5a;2  +  17a;  +  6  =  (5a;  +  2)(x  +  3). 
ox2  -  17 x  +  G  =  (5a;  -  2)(x  -  3). 
And,  since  5x3  —  2  x    1   =    13,  we  have  only  to  insert 
the  proper  signs  in  each  factor. 

In  (3)  the  positive  sign  must  predominate. 
In  (4)  the  negative  sign  must  predominate. 
Therefore  5  a;2  -f  13a;  —  6  =  (ox  —  2) (as  -f  3). 
ox2  -  13x  -  6  =  (5a;  -f  2)  (x  -  3). 
More  generally,  trinomials  of  the  form  ax2  +  bx  -f-  c,  (a 
not  a  square)    may  be   factored  more   readily   as   follows : 
Multiplying  by  a  we  get  a2x2  +  bax  +  ac.     Writing  z  for 
ax,  this  becomes  z2  -\-  bz  +  ac.      Factor  this  trinomial  by 
Art.  65,  replace  the  value  of  z,  and  divide  the  result  by  a. 
Thus, 

3.    Resolve  into  factors  6a;2  -  13a;  +  6. 
Multiplying  by  6  we  get  {&x)2  -  13  (6a;)  +  36. 
Putting  z  for  Gx  we  get 

z2  -  132  +  36, 
which,  being  factored,  gives 

(«-»)(« -4). 
Hence  the  required  factors  of  Gx2  —  13a;  -f-  6  are 

(2a;  -  3)(3a;  -  2). 

Ans.  (Sx  +  2)  (a;  +1), 

(2a;  +  l)(x  +  2). 
(2.i«  +  3)(a?  +  2). 
(3a;  +  2)(a:  +  2). 
(2a;  +  l)(a;  +  5). 


J (6a;  -  9)  (6a;  - 

-4) 

Resolve  into  factors 

4.    3a;2  +  5a;  +  2. 

5.    2a;2  +  5a;  +  2. 

6.    2a;2  -f  7x  +  6. 

7.    3a;"2  +  8a;  +  4. 

8.    2a;2  +  11a;  +  5. 

9. 

3x2  +  10.x-  +  3. 

If). 

3a;2  +  11a?  +  G- 

11. 

4a;2  +  Ha  -  3. 

Si. 

3a;2  +  x  -  2. 

13. 

2.x2  +  3a;  -  2. 

14. 

2x2  +  15*  —  8. 

15. 

3a;2  -  19a;  -  14. 

10. 

6x2  -  Six  +  35. 

17. 

3a;2  +  19^  -  14. 

18. 

4a;2  +  x  -  14. 

98       TO  FACTOR    TIIA'   DIFFERENCE    OF   TWO   SQUARES. 

Ans.  (ox  -f  l)(a;  +  3) 
(3a;  -f  2)  (x  +  3) 
(4a;  -  l)(a;  +  3) 
(3a;  _2)(*  +  l) 
(2,;  -  l)(x  +  2) 
(2x-  1)(»  +  8) 
(3a;  +  2)  (x  -  7) 
(3a;  -  5)  (2a;  -  7) 
(3a;  -  2)  (x  +  7) 
(4a;  -  7)  (a;  +  2) 

67.  To  Factor  an  Expression  which  is  the  Differ- 
ence of  Two  Squares.  —  From  (3)  of  Art.  41,  we  see 
that  the  difference  of  two  squares  is  equal  to  the  product  of 
the  sum  and  difference  of  their  square  roots.  Therefore, 
conversely,  to  find  the  factors  we  have  the  following 

Rule. 
Extract  the  square  roots  of  the  two  terms;  take  the  sum  of 
the  results  for  one  factor,  and  the  difference  for  the  other. 

EXAMPLES. 

1.    Resolve  into  factors  25a;2  —  1G?/2. 

25a;2  -  lGif  =  (5a;)2  -  (4y)2. 
Therefore  the  first  factor  is  the  sum  of  5a;  and  4?/,  and  the 
second  is  the  difference  of  5a;  and  4y. 

.-.     25a;2  -  lGif  =  (5a;  +  4?/)  (5a;  -  4y). 
The  intermediate  steps  may  usually  be  omitted. 
Resolve  into  factors 


2. 

a;2  -  9ft2. 

Ans.  (x  +  3a)  (as  —  3«). 

3. 

121  -  a;2. 

(11  +  »)(H  -  •'•)• 

4. 

y2  —  25a;2. 

(V  +  5a?)  (y  -  5x). 

5. 

36a;2  -  2562. 

((•„.  +  56)(6aj  -  56). 

6. 

xhf  -  36. 

(xy  +  G)(zy  -  6). 

7. 

ft-6-  -  \c2d\ 

(ah  +  2cd)(ab  -  2cd). 

TO   FACTOR    THE   DIFFERENCE    OF   TWO   S  ('FARES.       99 

8.    81a4  -  4:»./-4.  Ans.  (9a2  +  7a;2)  (9a2  -  7a;2) 

J    9a4  -  121.  (3a2  +  11)  (3a2  -  11) 

x*  -  25.  (t*  +  5)(aj»  -  5) 

x4a2  -  49.  (x-2a  +  7)  (x2a  -  7) 

a2  _  54^  (a  +  Sx3)  ^a  _  8aj3j 

68.  "When  One  or  Both  of  the  Squares  is  a 
Compound  Expression.  —  Here  the  same  method  is 
employed  as  in  the  last  Article. 

EXAMPLES. 

1.  Resolve  into  factors  (2a  -f  b)2  —  9a;2. 
The  sum  of  2a  +  b  and  3x  is  2a  +  b  -+■  3a;, 

and  their  difference  is  2a  -f-  o  —  3a;. 

.-.      (2a  +  b)'2  -  9x2  =  (2a  +  b  +  3a;)  (2a  +  6  -  3a;). 

If  the  factors  contain  like  terms  they  should  be  collected 
so  as  to  give  the  result  in  its  simplest  form. 

2.  Resolve  into  factors  (5a;  +  8y)2  —  (4a;  —  oy)'2. 
The  sum  of  5a;  -f-  8y  and  4a;  —  oy 

is  5a;  -f-  &J  +  4a;  —  3y  =  9a;  +     5y, 

and  their  difference  is 

5a;  +  Sy  -  4a;  +  Sy  =     a;  +  Uy. 
.-.      (5a;  +  8t/)2  -  (4a;  -  3y)2  =  (9a;  +  5y)  (a;  +  lly). 

Ans.  (a  +6  +  c)(o  +  6-  c) 

(a  -  6  +  c)  (a  —  b  —  c) 

(a-  +  ^  +  2^)  (a;  +  y  -  2z) 

(a?  +  2y  +  a)(o;  +  2y  -  a) 

(a  -  2a;  -f  /a)  (a  -  2x  -  b) 

(2x  -  3a  +  3c)  (2x  -  3a  -  3c) 

(2a;  +  y)y 

y(2x  -  y) 

(x  +  oy)  (x  +  ?/) 

(7a;  +  3)2  -  (5a;  -  4)2.  (12a;  -  l)(2a  +  7) 


Resolve 

!  into  factors 

o 
O. 

(« 

+  &)s  - 

-  C2. 

4. 

(a 

-  by  - 

-  c2. 

5. 

(a; 

+  yY  - 

-  4*2. 

C. 

(a; 

+  ?yY 

—  a2. 

7. 

(a 

-  2.i-)'2 

-  b2. 

8. 

(2a;  -  3a)2 

-  9c2 

iS< 

(X 

+  yY- 

-  a;2. 

w 

X2 

-Oj- 

a;)2. 

?y 

(X 

+  s;jy 

-  4/. 

100     TO  FACTOR   THE  DIFFERENCE   OF   TWO   SQUARES. 

69.  Compound  Quantities  expressed  as  the  Dif- 
ference of  Two  Squares.  —  Compound  expressions,  by 
suitably  grouping  the  terms,  can  often  be  expressed  as  the 
difference  of  two  squares,  and  so  be  resolved  into  factors. 

EXAMPLES. 

1 .  Resolve  into  factors  a2  +  2ax  +  x2  —  462. 
From  (1)  of  Art.  41,  a2  -f  2ax  +  x2  =  (a  +  %)*• 

.-.     a2  +  2ax  -f-  x2  -  462  =  (a  +  x)2  -  4b2, 
which  by  Art.  G8  =  (a  +  x  +  26)  (a  +  x  -  26). 

2.  Resolve  into  factors  9a2  —  c2  -f-  4cx  —  4se2. 
9aa  _  g2  +  4ca.  _  4a.a  _  9aa  _  (c2  _  Acx  _f_  ^ 

=  gaa  -  (c  -  2a;)2  by  (2)  of  Art.  41, 
—  (3a  +  c  —  2a?)  (3a  —  c  +  2x). 

3.  Resolve  into  factors  24xy  +  25  -  lG.r2  -  9?/2. 
24a;?/  +  25  -  IGor  -  dy2  =  25  -  (lG.r2  -  24z?/  +  9?/2) 

=  25-  (4a?  -  3?/) 2  by  (2)  of  Art.  41, 
=  (5  +  4x  -  3y)(Z  -  4x  +  3y). 

4.  Resolve  into  factors  2bd  -  c2  -  a2  +  «"2  4-  62  +  2ac. 
Here   we   see    that   the    expression    is   composed   of   two 

trinomials,  each  of  which  is  the  square  of  a  binomial  [(1) 

and  (2)  of  Art.  41]. 

.-.     2bd-c2-a2+d2+b2+2ac==b2+2bd  +  d2-(a2-2ac+c2) 

=  (b+d)2-(a-c)2 
=  (6-feH-a— c)  (6+d— a+c) . 


Resolve  into  factors 


5. 

x2  +  2xy  +  y2  -  a2. 

C. 

a?2  —  6aa>+9aa-16&a< 

7. 

4a2  +  [(ib  +  b2—  9c2. 

8. 

x2  +  a2  +  2<tx  —  y2. 

9. 

cfl   __   a;a   _   ?/2   +   2,-.V. 

10. 

Oa  +  y3+  2xy  -ix-y-. 

Ans.  (x  -f-  y  +  a)  (x  +  ?/  —  a) . 

(»  —  3rt  +  46)  (x  —  oa  —  46) . 

(2a  +  6  +  3c)  (2a  +  6  -  3c). 

(oj  +  a  +  y)  (x  -f  a  —  y). 

(c  +  •'•  -  ?/)(<;  -  a  +  y). 

(aj  +  y+2a#)(a:+y-2ajy), 


MISCELLANEOUS   CASES   OF  FACTORING.  10 i 

70.  To  Factor  an  Expression  which  can  be 
Written  as  the  Sum  or  the  Difference  of  Two 
Cubes.  —  Such  expressions  as  these  may  be  resolved  into 
factors  b}'  Art.  51. 

EXAMPLES. 

1.  Resolve  into  factors  8a;3  —  27y3. 

By  I.  of  Art.  51,  this  is  divisible  by  2x  —  Sy. 
.-.     8x3  -  27y3  =  (2a;)3  -  (3y)3 

=  (2x  -  3?/)  (4a;2  +  Gxy  +  9y*)x 
Note.  —  The  middle  term  Gxy  is  the  product  of  2x  and  oy. 

2.  Resolve  into  factors  343a;6  +  21  if. 

343a;6  -f  21yz  =  (7a;2  -f-  oy)  (49a;4  -  21x2y  +  9y*), 
(by  III.  of  Art.  51). 

Resolve  into  factors 

3.  8as  -  21if.  Ans.  (2a  -  Sy)(4a2  +  Gay  +  9>/2). 

4.  1  -  343a;3.  (1  -  7x)(\  +  7x  +  49a;2). 
(?)  asb3  +  512.  (ab  +  8)  (a2b2  -  Sab  +  G4). 

6\    343  +  8a;3.  (7  -f-  2a;)  (49  -  14a;  +  4x2). 

(T^21G.r3  -  343.  (6a;  -  7)  (36a;2  +  42a;  +  49). 

8.    27a;3  -  G-ly3.  (3a;  -  4?/)  (9a;2  +  12xy  +  1G//2). 

©  64a;6  +  125?/3.  (4a;2  +  5?/)  (16a;4  -  20x2y  +  25/). 

!J0,   216a;6  -  63.  (6a;2  -  b)  (36a;4  +  6a;26  +  b2). 

11.    «3  +  34363.  (a  +  76)  (a2  -  7a&  +  4962). 

71.  Miscellaneous  Cases  of  Resolution  into  Fac- 
tors. —  When  an  expression  can  be  arranged  as  the  differ- 
ence of  two  squares,  it  may  be  factored  either  by  (II.)  of 
Art.  51  or  by  (3)  of  Art.  41.  It  will  be  found  the  simplest, 
however,  first  to  factor  by  the  second  method,  using  the  rule 
for  factoring  the  difference  of  two  squares  (Art.  67). 


J  02  EXAMPLES. 

EXAMPLES. 

1.  Resolve  into  factors  16a4  —  8lb4. 

10a4  -  8164  =  (4a2  +  962)(4a2  -  9b2)  (Art.  67) 

=  (4a2  +  962)(2a  +  36)  (2a  -  3b)  (Art.  67). 

2.  Resolve  into  factors  a;6  —  y6. 

x6  -  yG  =  (a;3  +  f){pp  -  if)  (Art.  67) 

=  (»  +  #)  O2  -  ^  +  2/2)  (x  -  y)  (x<2  +  *9  +  2/2) 

(Art.  51,  I.  and  III.). 

The  student  should  be  careful  in  every  case  to  remove  all 
monomial  factors  that  are  common  to  each  term  of  an  ex- 
pression, and  place  them  outside  a  parenthesis,  as  explained 
in  Art.  63. 

3.  Resolve  into  factors  2Sx*y  -f  Gix3y  —  G0x2y. 
28x*y  +  64a%  -  G0x2y  =  Ax2y(7x2  -f-  16a;  -  15) 

=  ix2y{lx  -  5)(x  +  3)  (Art.  66). 

4.  Resolve  into  factors  x3a2  —  &y3a2  —  4x3b2  -f-  32ysb2. 
aft*2-  St/a2-  lx%2+  32/62=  «2(x'3-  8/)  -  W~(x3-  8y3) 

=  (x3-8f)(a2-W2) 

=  (x-2y)(x2+2xy+4f)(a  +  2h)(a-2b). 

5.  Resolve  into  factors  Ax2  —  25  ?/2  -|-  2a;  +  by. 

itf  _  95/  +  2a;  +  by  =  (2x  +  by)  (2x  -  by)  +  2a;  +  by 
=  (2a;  +  5?/)  (2a;-  5//+  1). 

Resolve  into  two  or  more  factors 

6.  a2  —  y2  —  2yz  —  z2.         Ans.  (a  +  y  +  z)(a  —  y  —  z) 

7.  Gx2  -  x  -  77.  (3a  -  11) (2*  +  7) 

8.  a;c-4096.       (a?+4)  (a;2-4a;-f-16)  (a;-4)  (a;2 + 4a; +16) 

9.  a;2  —  a2  -f  y2  —  2a,'?/.  (a;  —  y  +  a)  (a;  —  y  —  a) 

10.  aca;2  —  hex  -j-  aefcc  —  bd.  (ex  +  tf)(a»  —  6) 

11.  (a  +  6  +  c)2  -  (a  -  6  -  c)"-  4a(6  +  c) 
Other  expressions  which,  by  a  slight  modification,  can  be 

arranged  as  the  difference  of  two  squares,  may  be  factored 
by  Art.  67. 


EXAMPLES.  103 

12.  Resolve  into  factors  xA  +  x-f  +  y4- 

x*  +  xy  -f  2/4  =  a;4  +  2ajy  +  2/4  -  a;'2?/2 

=  (^2  +  2/2)2  -  xy 

=   (^2  +  2/2  +  JgO(rf  +  ya-ay). 

13.  Resolve  into  factors  xi  —  lox2f  -f-  9?/4. 
£4  -  15a?y  +  %4  =  (z2  -  3?/2)2  -  9afy2 

=  (x*  -  3?/2  +  3xy)(x*  -  Stf  -  Sxy). 

Expressions  which  can  be  put  into  the  form  x8  ±  -  may 

y3 

be  factored  by  the  rules  for  resolving  the  sum  or  the  differ- 
ence of  two  cubes  (Art.  70). 

o 

14.  Resolve  iuto  facto rs 27V5. 

of 

I  -■lf  -(;)'-  (W' 


-e-*xs+?+",> 


O,  2  Q 

15.    Resolve  a2^3 x3  +  -^  into  four  factors. 

2/3  Z/3 

aV-%2-^3+  §-W(«2- 1)  -4(«'2-l) 
i/3  */3  2/3 

=  ( 


a2-l)^-i) 


16.  Resolve  a9  —  G4a3  —  a6  -j-  64  into  six  factors. 
The  expression 

=  a3(a6-64)-(«6-64) 

=  (a6-64)(a3-l) 

=  (a3+8)(a3-8)(«3-l) 

=  (a  +  2)(a2-2«+4)(a^2)(a2+2«+4)(a-l)(a2-f-a  +  l). 
Resolve  into  factors 

17.  a**+16a?  +  256.         -4w*.  (^+43+16)  (x*-4x+ 16). 

18.  a*  +  f  -  7«y.         (z2  +  Sxy  +  y2)  (x2  -  Sxy  +  if) . 

19.  8la*+9a*b*+b*.         (9a2+  3o6  +  &2)  (9a2  -  Sab  +  ft2). 

20.  a**—  19aJ2y2+25y4.         (^2+  3xy—  o^/2)  (s2—  3a;#—  5#2) . 


104  EXAMTLES. 

By  a  skilful  use  of  factors,  the  actual  processes  of  multi- 
plication aud  division  can  often  be  partially  or  wholly 
avoided. 

21.  Multiply  2a  +  36  -  c  by  2a  -  3b  +  c. 

The  product  =  [2a  +  (3b  -  c)][2a  -  (3b  -  c)] 
=  (2a)2  -  (3b  -  c)2  [(3)  of  Art.  41] 
=  4a2  -  9b2  +  Mc  -  <S\ 

22.  Divide  the  product  of  a,*2  —  bxy  +  Gy2  and  a;  —  4y 
by  x2  —  Ixy  +  I2y2. 

We  might  multiply  the  first  two  expressions  together  and 
then  divide  the  result  by  the  third.  But  by  factoring  the 
first  and  third  expressions,  and  denoting  the  division  by 
means  of  a  fraction  (see  Art.  78) ,  the  work  will  be  much 
shorter. 

Thus,  the  required  quotient 

=  (a*2  -  5xy  +  6y2)  (x  -  Ay) 

x2  —  Ixy  +  I2y2 
_  (x  -  3?/)  (x  -  2?/)  (x  -  4?/) 
(x  -  Ay)  (x  -  3y) 

=  x  -  2y. 

23.  Divide  the  product  of  2a,*2  4-  x  —  6  and  Gar"  —  5a*  4-  1 
by  3x2  +  5a*  -  2.  4»s.  (2a;  -  3)  (2a;  -  1). 

Find  the  product  of 

24.  2as—  7y+3z  and  2a?-f  7y—Sz.       4x2—A9y2+A2yz—dz2. 

25.  a*3+2a*2?/  +  2a*?/2+?/3  and  a*3 -2x2y  +  2 xy2-y3.        xc'-yc'. 
2G.    a*3  -4a;2  +  8a;  -  8  and  a;3  +  4ar  +  8a;  +  8.         a*G  -  04. 

Divide 

27.  2a*(a*2-l)(a*-f-2)  by  a*2  +  a*-2.  2a>(aJ+l). 

28.  (x2  +  7a?  +  10)  (a;  +  3)  by  a;2  +  5a*  +  6.         a;  +  5. 

29.  5a*(a*-ll)(a*2-a*-15G)  by  a^f-aP— 18S*.     5(a*-13). 

30.  a9  -  b°  by  (a2  +  ab  +  h-)  (a8  +  a868  +  &e)  •         a  -  b. 

31.  [or  +  (a  -  &)a*  -  «&]  [a;8  -  (a  -  b)x  -  ah]  by 
a2  4-  (a  -f-  fr)a*  +  a&.  Jj/.s.  (x  —  a)(x  —  b). 


GREATEST   COMMON   DIVISOR  —  DEFINITIONS.        105 

GREATEST    COMMON    DIVISOR. 

72.  Definitions.  —  A  Common  Divisor  of  two  or  more 
expressions  is  an  expression  that  will  divide  each  of  them 
exactly. 

Hence,  every  factor  common  to  tico  or  more  expressions  is 
a  common  divisor  of  those  expressions  (Art.  62). 

Thus,  in  4«'2&,  G«362,  and  a463,  a2  occurs  as  a  factor  of  each 
quantity  ;  b  also  occurs  as  a  factor  of  each  quantity  ;  a2  and 
b  are  therefore  common  divisors  of  these  three  quantities. 

The  Greatest  Common  Divisor  of  two  or  more  Algebraic 
expressions  is  the  expression  of  highest  degree  (Art.  18) 
which  will  divide  each  of  them  exactly. 

Note.  —  The  term  greatest  common  divisor,  which  has  been  adopted 
from  Arithmetic,  does  not  imply  in  Algebra  that  it  is  numerically  the 
greatest,  but  that  it  is  the  factor  of  greatest  degree.  The  student  is 
cautioned  against  being  misled  by  the  analogy  between  the  Algebraic 
and  the  Arithmetic  greatest  common  divisor.  He  should  notice  that 
no  mention  is  made  of  numerical  magnitude  in  the  definition  of  the 
Algebraic  greatest  common  divisor.  In  Arithmetic,  the  greatest 
common  divisor  of  two  or  more  whole  numbers  is  the  greatest  whole 
number  which  will  exactly  divide  each  of  them.  But  in  Algebra,  the 
terms  greater  and  less  are  seldom  applicable  to  those  expressions  in 
which  definite  numerical  values  have  not  been  assigned  to  the  various 
letters  which  occur.  Besides,  it  is  not  always  true  that  the  Arith- 
metic greatest  common  divisor  of  the  values  of  two  given  expressions 
obtained  by  assigning  any  particular  values  to  the  letters  of  those 
expressions,  is  the  numerical  value  of  the  Algebraic  greatest  common 
divisor  when  those  same  values  of  the  letters  are  substituted  therein, 
as  will  be  shown  later  (Art.  74).  For  this  reason,  some  writers  have 
used  the  terms,  highest  common  divisor,  and  highest  common  factor, 
instead  of  the  term  greatest  common  divisor.  But  to  avoid  employing 
a  new  phrase,  and  in  conformity  with  well-established  usage,  wTe  shall 
retain  the  old  term  greatest  common  divisor. 

The  abbreviation  G.  C.  D.  will  often  be  used  for  shortness  instead 
of  the  words  greatest  common  divisor. 

73.  The  Greatest  Common  Divisor  of  Monomials, 
and  of  Polynomials  which  can  be  easily  Factored. 

—  Let  it  be  required  to  find  the  greatest  common  divisor  of 
21aVy,  35a2x*y,  28a3x2y\  andl4a5*V2. 


106        GREATEST   COMMON    DIVISOR    OF   MONOMIALS. 

By  separating  each  expression  into  its  prime  factors,  we 
have  2la4a?y    =  7  x  daaaaxxxy, 

3oa2x4y    =  7  x  baaxxxxy. 
28a3x-yi  =  7  x  2  x  2aaaxxyyyy. 
l±abx2y2  =  7  X  2aaaaaxxyy. 
By  examining  these  expressions  we  find  that  7,  aa,  xx, 
:md  //  are  the  only  factors  common  to  all  of  them.     Hence 
all  the  expressions  can  be  exactly  divided  by  either  of  these 
factors,  or  by  their  product,  7a2x2y,  which  is  therefore  their 
greatest  common  divisor. 

Find  the  G.  C.  D.  of  4cx3  and  Vex*  +  4c2x\ 
Resolving  each  expression  into  its  factors,  we  have 
4cx3  =  2cx2  x  2x. 
2cx3  +  4c2x2  =  2cx2(x  -f  2c). 
Here  it  is  clear  that  both  expressions  are  divisible  (1)  by 
2,  which  is  the  numerical  greatest  common  divisor  of   the 
coefficients,  (2)  by  c,  and  (3)  by  x2. 

.-.     G.  C.  D.  =  2cx2. 
Find  the  G.  C.  D.  of  3a2  +  dab,  a3  -  dab2,  a3  4  Ga2b  +  dab2. 
Resolving  each  expression  into  its  factors,  we  have 
3a2  +  dab  =  3a  (a  +  36). 

a3  +  0a26  4-  9a&2  =  a(a  4-  3d) (a  +-36). 

.-.     G.  C.  D.  =  a(a  4-  36). 

.Find  the  G.  C.  D.  of  «{a  -  x)\  ax  {a  -  a)8,  2ax{a  -  x)\ 

Resolving  into  factors,  we  have 

x(a  —  x)2  =  x(a  —  x){a  —  x). 

ax(a  —  x)3  =  ax  (a  —  x)(a  —  x)(a  —  x). 

2ax(a  —  xy  =  2ax(a  —  x)(a  —  x)(a  —  x)(a  —  x). 

.-.     G.  C.  D.  =  x(a  -  x)2. 

Hence  the  following 

Rule. 

Resolve  each  expression  into  its  prime  factors,  and  take  the 
product  of  all  the  factors  common  to  all  the  expressions, 
giving  to  each  factor  the  highest  power  which  is  common  to  all 
the  given  expressions. 


GREATEST   COMMON   DIVISOR    OE  POLYNOMIALS.     107 

EXAMPLES. 

Find  the  G.  C.  D.  of 

1.  4ab\  2a?b,  Gab3.  Arts.  2ab. 

2.  3ay,  x3y2,  x2y3.  xY- 

3.  6xy2z,  8x2y3z2,  4xyz2.  2xyz. 

4.  5a3b\  loabc2,  10a2b2c.  Mb. 

5.  9x2y2z2,  12xyh,  6x3y2z3.  Zxy2z. 
0.  Sa%  Qabxy,  I0abx3y2.  2ax. 

7.  a2  -f  ab,  a2  —  b2.  a  +  6. 

8.  (x  -f  ?/)-2,  ar  -  ?/2.  a  +  ?/. 

9.  a3  +  x2y.  x3  -f-  ?/3.  #  -f  ?/. 

10.  a3  —  a?x,  a3  —  ax2,  a4  —  ax3.  a  (a  —  x). 

11.  a;4  -  27asas,  (a;  -  3a)2.  a;  -  3a. 

12.  a#  —  ?/,  x*y  —  xy.  y(x  —  1). 

13.  ax2  +  2a23  +  a3,  2az2  -  4a2z  -  Ga3,  S(ax  +  «2)2. 

yl?is.  a  (as  +  a). 

74.  The  Greatest  Common  Divisor  of  Expressions 
that  cannot  be  Readily  Resolved  into  Factors. — 
To  find  the  G.  C.  D.  in  such  cases,  we  adopt  a  method 
analogous  to  that  used  in  Arithmetic  for  finding  the  G.  C.  D. 
of  two  or  more  numbers. 

The  method  depends  on  two  principles. 

1.  If  an  expression  contain  a  certain  factor,  any  multiple 
of  that  expression  is  divisible  by  that  factor. 

Thus,  if  F  divides  A  it  will  also  divide  mA.  For  let  a 
denote  the  quotient  when  A  is  divided  by  F;  then  A  =  aF '; 
therefore  mA  =  maF ;  and  therefore  F  divides  mA. 

2.  If  two  expressions  have  a  common  factor,  it  ivill  divide 
their  sum  and  their  difference;  and  also  the  sum  and  the 
difference  of  any  midtiple  of  them. 

Thus,  if  F  divides  A  and  B,  it  will  divide  mA  ±  nB.  For 
since  F  divides  A  and  B,  we  may  suppose  A =aF^  and  B=bF\ 
therefore  mA  ±  nB  =  maF  ±  vbF 

=  F(ma  ±  nb). 
Therefore  F  divides  mA  ±  nB. 


108     GREATEST   COMMON   DIVISOR    OF  POLYNOMIALS. 

We  can  now  prove  the  rule  for  finding  the  G.  C.  D.  of  any 
two  compound  Algebraic  expressions. 

Let  A  and  B  denote  the  two  expressions.  Let  them  be 
arranged  in  ascending  or  descending  powers  of  some  common 
letter ;  and  let  the  highest  power  of  that  letter  in  B  be  either 
equal  to  or  greater  than  the  highest  power  in  A. 

Divide  B  by  A  ;  let  p  be  the  quotient  and  C  the  remainder. 
Suppose  C  to  have  a  simple  factor  m.  Remove  this  factor, 
and  so  obtain  a  new  divisor  D.  Suppose  further,  that  in 
order  to  make  A  divisible  by  D  it  is  necessary  to  multiply  A 
by  a  simple  factor  n.  Divide  nA  by  D  ;  let  q  be  the  next 
quotient,  and  E  the  remainder.  Divide  D  by  E ;  let  r  be 
the  quotient,  and  suppose  that  there  is  no  remainder.  Then 
E  will  be  the  G.  C.  D.  required. 

The  operation  of  division  will  stand  thus : 

A)B(P 
pA 

D)nA(q 
qD 

E)D(r 
rE 

First,  to  show  that  E  is  a  common  divisor  of  A  and  B. 
From  the  above  division  we  have  the  following  results : 

D  =  rE. 

nA  =  qD  -f  E  =  qrE  +  E      =  (qr  +  1)E. 

B  =  pA  +  C  =  pA  +  mD  =  MrE  +  pE  +  mrJB 

=  \ — —    +  mr)E- 

Therefore  E  is  a  common  divisor  of  A  and  B. 

Second,  to  show  that  E  is  the  greatest  common  divisor  of 
A  and  B. 


EXAMPLES.  109 

By  (2)  of  this  Art.  every  common  factor  of  A  and  B 
divides  also  B  —  pA,  that  is  C,  and  therefore  D  (^ince  m  is 
a  simple  factor) .  Similarly  as  it  divides  A  and  D  it  divides 
nA  —  qD,  that  is  E.  But  no  expression  of  higher  degree 
than  E  can  divide  E.  Therefore  E  is  the  greatest  common 
divisor  of  A  and  B. 

The  greatest  common  divisor  of  three  expressions,  A,  B< 
(7,  may  be  obtained  as  follows : 

First  find  D,  the  G.  C.  D.  of  any  two  of  them,  say  of  A 
and  B;  next  find  F,  the  G.  C.  D.  of  D  and  C;  then  F  will 
be  the  G.  C.  D.  of  A,  B,  C.  For  D  contains  every  factor 
which  is  common  to  A  and  B  (Art.  72);  and  as  F  is  the 
G.  C.  D.  of  D  and  C,  it  contains  every  factor  common  to  D 
and  (7,  and  therefore  every  factor  common  to  A,  B,  and  (7. 
Hence  F  is  the  G.  C.  D.  of  A,  B,  C. 

EXAMPLES. 

1.    Find  the  G.  C.  D.  of  x2-±x  +  3  and  4xs-  9a;2-15a;+18. 
x2  -  4x  +  3)4:c3  -     9x2  -  15a;  +  18(4aJ  +  7 
4a3  -  16s2  +  12a? 

7x2  -  21x  +  18 
7x2  -  28a;  +  21 


x  -    3)  x2  -  Ax  +  3 (a;  -  1 
a2  -  3a; 


-  x  +  3 

—  a;  +  3 


Therefore  the  G.  C.  D.  is  x  -  3. 


Explanation.  —  First  arrange  the  given  expressions  according  to 
descending  powers  of  x.  Take  for  dividend  that  expression  whose 
first  term  is  of  the  higher  degree;  and  continue  each  division  until 
the  first  terra  of  the  remainder  is  of  a  lower  degree  than  the  first  term 
of  the  divisor.  When  the  first  remainder,  x  —  3,  is  made  the  divisor, 
we  put  the  first  divisor  to  the  right  of  it  for  a  dividend,  and  after 
obtaining  the  new  quotient,  x  —  1,  we  have  nothing  for  a  remainder. 
Hence,  as  in  Arithmetic,  the  last  divisor,  x  —  3,  is  the  G.  C.  D. 
required. 


110 


EXAMPLES. 


2.    Find   the   G.  C.  D.  of   8a;3  -   2a;2  -  53a;  -  39    and 


4^3  _  3x2  _  2±x 


(J. 


4a;3  - 

-  3a;2  -  24a;  -  9 

8a;3  -  2a;2  -  53a;  - 

39 

4a;3  - 

-  5a;2  -  21a; 

8a;3  -  Gx2  -  48a;  - 

18 

2a;2  -    3a;  -  9 

4a;2  —     5a;  — 

21 

2a;2  -    6a; 

4a;2  -     6a;  - 

18 

Sx  -  9 

x  — 

3 

3a;  -  9 

2a; 


Therefore  the  G.  C.  D.  is  x  -  3. 

Explanation.  —  First  arrange  the  given  expressions  according  to 
descending  powers  of  x.  The  expressions  so  arranged  having  their 
first  terms  of  the  same  order,  we  take  for  divisor  that  whose  highest 
power  has  the  smaller  coefficient,  and  arrange  the  work  in  parallel 
columns,  as  ahove.  (1)  At  the  first  division  we  put  the  quotient  2  to 
the  right  of  the  dividend.  (2)  When  the  first  remainder  4.<-  —  5a  —  21 
is  made  the  divisor  we  put  the  quotient  x  to  the  left  of  the  dividend. 
(3)  When  the  second  remainder  2x2  —  Sx  —  9  is  made  the  divisor  we 
put  the  quotient  2  to  the  right  of  the  dividend.  (4)  When  the  third 
remainder  x  —  3  is  made  the  divisor  we  put  the  quotients  2x  and  3  to 
the  left  of  the  dividend,  and  so  on. 

This  method  is  used  only  to  determine  the  compound  factor 
of  the  G.  C.  D.  Simple  factors  of  the  given  expressions 
must  first  be  separated  from  them,  and  the  G.  C.  D.  of  these, 
if  they  have  any,  must  be  reserved  and  multiplied  into  the 
compound  factor  obtained  by  the  rule. 

3.  Find  the  G.  C.  D.  of  6a;4  -  26a;3  +  46a;2  -  42a;  and 
18a;4  -f-  3a;3  -  132a;2  +  63a;. 

We  have 

6a;4-26a;8+  46a;2-42a;=2a;(3a;3-13a;2+23a;-21),     .   (1) 

and 

18a;4 +  3a;3-132a;2  +  63a;=3a;(6a;3-f     a;2 -44a; +21).     .    (•_>) 

The  simple  factor  2  is  found  in  the  first  expression  and 
not  in  the  second;  therefore  it  forms  no  part  of  the  G.  C. 
D.,  and  may  be  rejected.  Likewise  the  simple  factor  3, 
occurring  in  the  second  expression  and  not  in  the  first,  may 


EXAMPLES. 


Ill 


be  rejected  as  forming  no  part  of  the  G.  C.  D.  But  the 
simple  factor  x  is  common  to  both  expressions,  and  is 
therefore  a  factor  of  the  G.  C.  D.  and  must  be  reserved. 
Rejecting  therefore  the  simple  factors  2  and  3  as  forming  no 
part  of  the  G.  C.  D.,  and  reserving  the  common  factor  a;  as 
forming  a  part  of  the  G.  C.  D.,  and  arranging  in  parallel 


3a;3  —  13ar  -f  23a; 


21 


Oa;3  +       x2  -  Ux  +  21 


6a;3  -  26a;2  +  46a; 


42 


27a;2  -  90a;  +  63 

The  first  division  ends  here,  since  27a;2  is  of  a  lower  degree 
than  3a;3.  If  we  now  make  27a;2  —  90a;  -f-  63  a  divisor  we 
find  that  it  is  not  contained  in  3a:3  —  13a;2  +  23a;  —  21  with 
an  integral  quotient.  But,  noticing  that  27a;2  —  90a;  -f-  63 
may  be  written  in  the  form  9(3ar  —  10a;  -+-  7),  and  remem- 
bering that  the  G.  C.  D.  we  are  seeking  is  contained  in  the 
remainder  9  (3a;2  —  10a;  +  7),  and  that,  since  the  two  expres- 
sions 3a;3  —  13a;2  -f  23a;  —  21  and  Gx3  +  x'2  —  44a;  +21 
have  no  simple  factors,  therefore  their  G.  C.  D.  can  have 
none,  we  conclude  that  the  G.  C.  D.  must  be  contained  in 
the  factor  3a;2  —  10a;  +  7,  and  that  therefore  we  can  reject 
the  simple  factor  9,  and  go  on  with  the  divisor  3a;2  —  10a;  -f-  7. 
Resuming  the  work,  we  have 

x    Sx3  -  13a;2  +  23a;  -  21 
3a;3  -  10a;2  +     7x 

21 


-1 


-  3a;2  +  16a; 

-  3ar  -f  10a; 


2)  6a; 


14 


3a;2 
3a;2 

— 

10a;  + 
7x 

7 

— 

ox  + 
3a;  + 

7 
7 

-1 


3a;  —     7 
Therefore  the  G.  C.  D  is  x(3x  -  7). 

The   factor  2  was   removed   for  the  same  reason  as  the 
factor  9. 


4.    Find    the    G.  C.  D.  of   2a;3 
3a;3  -  2or  +  x  -  2  (2). 


+  ar  —   x 


_   2 


2   (1)    and 


112 


EXAMPLES. 


As  the  expressions  stand  neither  can  be  divided  by  the 
other  without  obtaining  a  fractional  quotient.  This  difficulty 
cannot  be  obviated  by  removing  a  simple  factor, 'since  neither 
expression  contains  a  simple  factor.  We  may  however  intro- 
duce a  suitable  factor  into  either  expression,  just  as  in  Ex.  3 
we  removed  a  factor  when  we  could  no  longer  proceed  with 
the  division  without  a  fractional  quotient.  The  given  expres- 
sions (1)  and  (2)  have  no  common  simjrte  factor,  therefore 
their  G.  C.  D.  can  have  no  simple  factor,  and  hence  cannot 
be  affected  if  we  multiply  either  of  them  by  any  simple 
factor. 

Multiplying  (2)  by  2  and  taking  it  for  dividend,  we  have 


2x3  +  x2  - 

7 

a;  — 

2 

6a;3-   4a;2  +  2x-    4 
6x*+      3a;2-  3a;-  6 

-2a; 

14a;3  +  7a;2- 
14a;3-  10a;2- 

7a;  — 
4a; 

14 

—   7ar  +  5a; -j-  2 
17 

17a; 

17a;2- 

17a;2- 

3x  - 
17a; 

14 

-119a;2  +  85a; +  34 
-  119a;2  +  21a; +  98 

14 

14a;- 
14a;- 

14 
14 

64)  64a; -64 
x-    1 

Therefore  the  G.  C.  D.  is  x  -  1. 

In  this  example,   after  the  first  division  the   factor  7  is 
introduced  because  the  first  remainder  —7a;2  +  5a;  +  2  will 


not  divide  the  first  divisor  2a;3  +  x1 


2. 


After  the  second  division  the  factor  17  is  introduced 
because  the  second  remainder  17a;2  —  3a;  —  14  will  not 
divide  the  second  divisor  —7a;2  +  5x  +  2.  Finally  the 
factor  64  is  removed  as  explained  in  Ex.  3. 

Note.  —  The  difference  between  the  Algebraic  G.  C.  D.  and  the 
Arithmetic  G.  G.  D.  can  be  seen  by  an  example. 

Factor  (1)  and  (2)  of  last  example  as  follows: 

2a;3  +    a;2  -  a;  -  2  =  (x  -  l)(2.r2  +  3a  +  2), 
and    3a;3  -  2xz  +  x  -  2  =  (x  -  l)(3ar  +    x .+  2). 


EXAMPLES.  113 

Now  since  the  G.  C.  D.  of  these  expressions  is  x  —  1, 
the  factors  2x*  -j-  3x  -f  2,  and  ox2  +-  x  -f-  2,  have  no  common 
factor.     But  if  we  put  x  =  4,  then 

2aj»  +    x2  -  a  -  2  =  138, 
and  3x*  -  2x2  -f  *  -  2  =  162, 

and  the  G.  C.  D.  of  138  and  162  is  6,  while  3  is  the 
numerical  value  of  the  Algebraic  G.  C.  D.,  x  —  1.  Thus 
the  numerical  value  of  the  Algebraic  G.  C.  D.  does  not  agree 
with  the  numerical  value  of  the  Arithmetic  G.  C.  D. 

The  reason  may  be  explained  as  follows  ;  the  expressions 
2.r2  +  3x  -f-  2  and  3x2  -f-  x  +  2  have  no  Algebraic  common 
factor ;  but  when  x  =  4  the}'  become  equal  to  46  and  54 
respectively,  and  therefore  have  a  common  Arithmetic  factor 
2,  which,  multiplied  into  x  —  1  or  3,  gives  6  for  the  nu- 
merical value  of  the  Arithmetic  G.  C.  D.,  while  3  is  the 
numerical  value  of  the  Algebraic  G.  C.  D.  In  the  same  way 
it  may  be  shown  that  if  we  give  particular  numerical  values 
to  the  letters  in  any  two  expressions,  and  in  their  Algebraic 
G.  C.  D.,  the  numerical  value  of  the  G.  C.  D.  is  by  no 
means  necessarily  the  Arithmetic  G.  C.  D.  of  the  values  of 
the  expressions. 

We  may  now  enunciate  the  rule  for  finding  the  greatest 
common  divisor  of  two  compound  Algebraic  expressions. 

Rule. 

Arrange  the  given  expressions  according  to  the  descending 
powers  of  the  same  letter.  Divide  that  expression  ivhich  is 
of  the  higher  degree  by  the  other ;  or,  if  both  are  of  the  same 
degree,  divide  that  ichose  first  term  has  the  larger  coefficient 
by  the  other;  and  if  there  is  no  remainder  the  first  divisor 
will  be  the  required  greatest  common  divisor. 

If  there  is  a  remainder  divide  the  first  divisor  by  it,  and 
continue  thus  to  divide  the  last  divisor  by  the  last  remainder, 
until  a  divisor  is  obtained  which  leaves  no  remainder;  the 
last  divisor  ivill  be  the  greatest  common  divisor  required. 


114        LEAST   COMMON   MULTIPLE — DEFINITIONS. 

Note  1.  —  Before  beginning  the  division,  all  simple  factors  of  the 
given  expressions  must  be  removed  from  them,  and  the  greatest 
common  divisor  of  these  must  be  reserved  as  a  factor  of  the  G.  C.  D. 
required.     (See  Ex.  3.) 

Note  2.  —  Either  of  the  given  expressions  or  any  of  the  remain- 
ders may  be  multiplied  or  divided  by  any  factor  which  does  not  divide 
both  of  the  given  expressions.     (See  Ex.  4.) 

Note  3. — Each  division  must  be  continued  until  the  remainder  is 
of  a  lower  degree  than  the  divisor. 

Find  the  G.  C.  D.  of 

5.  a3  +  2a;2  -  13a  +  10,  and  a3  +  x2  -  10a  +  8. 

Ans.  x2  —  3a  -f-  2. 

6.  a3-5a2-99a+40,  and  a3-0a2-8Ga -f  35.    x'-13x+5. 

7.  a3-a2-5a-3,  and  a3- 4a2- 11a -G.         a2-f-2a-fl. 

8.  a3  +  3a2  -  8a  -  24,  and  a3  +  3a2  -  3a  -  9.         a  +  3. 

9.  2a3+4a2-7a-14,andCa3-10a2-21a+35.        2a2-7. 

LEAST    COMMON    MULTIPLE. 

75.  Definitions.  —  A  Multiple  of  an  expression  is  any 
expression  that  can  be  divided  by  it  exactly. 

Hence,  a  multiple  of  an  expression  must  contain  all  the 
factors  of  that  expression.     Thus, 

Qa2b  is  a  multiple  of  3  or  2  or  G  or  a  or  b. 

A  Common  Multiple  of  two  or  more  expressions  is  an 
expression  that  can  be  divided  by  each  of  them  exactly ; 
or,  it  is  one  of  which  all  the  given  expressions  are  factors. 

Thus,  the  expression  ab2c3  is  a  common  multiple  of  the 
expressions,  a,  6,  c,  ab,  a&c,  ab2,  &2c3,  etc.,  or  of  the  expres- 
sion itself ;  but  it  is  not  a  multiple  of  «2,  nor  of  Z>3,  nor  of 
any  symbol  which  does  not  enter  into  it  as  a  factor. 

The  Least  Common  Multiple  *  of  two  or  more  Algebraic 
expressions  is  the  expression  of  least  degree  which  is  divisi- 
ble by  each  of  them  exactly. 

*  Called  also  lowest  common  multiple.  The  term,  leant  common  multiple,  Ifl 
objected  to  by  some,  for  a  reason  similar  to  tbo  one  for  wbieb  tbey  object  to  tbo 
term  greatest  common  divisor. 


LEAST   COMMON   MULTIPLE    OF   MONOMIALS.  115 

Hence,  the  least  common  multiple  of  two  or  more  expres- 
sions is  the  product  of  all  the  factors  of  the  expressions,  each 
factor  being  taken  the  greatest  number  of  times  it  occurs  in 
any  of  the  expressions. 

The  abbreviation  L.  C.  M.  is  often  used  instead  of  the 
words  least  common  multiple. 

Xote.  —  Two  or  more  expressions  can  have  only  one  least  common 
multiple,  while  they  have  an  indefinite  number  of  common  multiples. 

76.  The  Least  Common  Multiple  of  Monomials, 
and  of  Polynomials  which  can  be  easily  Factored. 

—  Let  it  be  required  to  find  the  least  common  multiple  of 
21aVy,  35a2a42/,  28a3x2y\  and  Ua\v2y2. 

By  separating  each  expression  into  its  prime  factors,  we 
have  2\ah?y    =  3    x  7a\^y, 


Shcrrfy    =  5 

X  7a2x4y, 

28a%y  =  22 

X  7asajV, 

14a5afy2  =  2 

X  7a5x2y2. 

,  the  L.  C.  M.  =  7 

X  3  x  5 

x  2WbY 

=  420rt5;/;y ; 
for  420  is  the  numerical  L.  C.  M.  of  the  coefficients ;  a5  is 
the  lowest  power  of  a  that  is  divisible  by  each  of  the  quan- 
tities a4,  a2,  a3,  a5 ;  x*  is  the  lowest  power  of  x  that  is 
divisible  by  each  of  the  quantities  x8,  x4,  x2 ;  and  y*  is  the 
lowest  power  of  y  that  is  divisible  by  each  of  the  quantities 

2.    Find  the  L.  C.  M.  of  §x*(a  -  x)2,  8a2(a  -  x)3,  and 
12a2x2(a  -  x)\ 
Resolving  into  factors,  we  have 

6a?  (a  -  x)2  =  3  x  2a?(a  -  x)2, 
8a\a  -  x)3  =  2  x  2  x  2a2(a  -  x)3, 
12a2z2(a  -  xy  =  3  x  2  x  2a2x2{a  -  x)K 
Hence,  the  L.  C.  M.  =  3  x  28a*Bs(a  -  xy 
=  24aV(a  -  a;)4. 
For  it  consists  of  the  product  of  (1)  the  numerical  L.  C.  M. 
of  the  coefficients,  and  (2)  the  lowest  power  of  each  factor 


116  EXAMPLES. 

which  is  divisible  by  every  power  of  that  factor  occurring  in 
the  given  expressions. 

3.    Find   the    L.    C.    M.    of    3a2   +    9ab,    2a3   -    18c*2, 
a»  +  Qa2b  +  9a62,  as  +  5a26  -f  6a62. 
Resolving  into  factors,  we  have 

3a2  4-    9ab    =  3a  (a  +  36), 
2a3  -  18a62  =  2a(a  +  36)  (a  -  36), 
a3  +  6a26  4-     9a62  =     a(a  +  36)2, 
a3  +  oa26  +     6a62  =     a(a  4-  36)  (a  +  26). 
Hence  the  L.  C.  M.  =  6a(a  +  36)2(a  -  36)  (a  +  26) 
Hence  the  following 

Rule. 

Resolve  each  expression  into  its  prime  factors,  and  take 
the  product  of  all  the  factors,  giving  to  each  factor  the  highest 
exponent  ivhich  it  has  in  the  given  expressions. 

If  the  expressions  are  prime  to  each  other,  their  product 
is  the  least  common  multiple. 

EXAMPLES. 


Find  the  least  common  multiple 

of 

1.    5a26c3,  4a62c. 

^ws.  20aW. 

2.    12a6,  Sxy. 

24ab.nj. 

3.    2a6,  36c,  4ca. 

\2abc. 

4.    a26c,  62ca,  c2«6. 

aW. 

5.    5a2c,  Gc62,  36c2. 

30aW. 

6»    x2,  x2  —  3x. 

a:2(z  -  3). 

7.    21a3,  7x2(x  +1). 

21x8(s  4-  1). 

8.    a2  +  «6,  ab  +  62. 

ab(a  4-  6). 

9.    Gx2  -  2x,  9.x2  -  3x. 

6ac(3aj  -  1 ) . 

10.    x2  +  2s,  x2  -f  3x  +  2. 

.<•(./• 

4-  2)(x  4-  1). 

11.    x2  +  4a;  +  4,  a2  +  5.x  +  ( 

(•'• 

4-  2)\x  4-  3). 

12.    x2  -f  x  -  20,  x2  -  10x  + 

2  1 ,  ./■-' 

—   X 

-  80. 

^l//.s. 

(•'■  + 

5)  (SB 

-4)(x-6). 

LEAST    COMMON    MULTIPLE    OF  POLYNOMIALS.       117 


X 

2a;4 -f  a;3- 20a;2—  7a; +24 
2x4-\-7xs           -9a; 

2a;4-h3a?}-13a;2-7a;+15 
2jb*+  x3  -20a;2-  7a; +24 
2a;3 +   7a;2         -   9 
2a;3  +   4a;2- Ga; 

3 

-Ga;3-20a;2+2a;+24 
-Ga;3-21a;2         +27 

a;2+2a;-   3 

3ar+Ga;—   9 
3a?+6x-  9 

77.  When  the  Given  Expressions  cannot  be 
Resolved  into  Factors  by  Inspection.  —  To  find  the 
least  common  multiple  of  two  compound  Algebraic  expres- 
sions in  such  cases,  the  expressions  must  be  resolved  by 
finding  their  G.  C.  D. 

1.  Find  the  L.  C.  M.  of  2x4  +  x3  -  20a;2  -  7a;  +  24  and 
-_V4  +  3a;3  -  13a;2  -  7x  +  15. 

We  first  find  their  G.  C.  D. 

1 


2x 


.-.     G.  C.  D.  =  a;2  +  2a;  -  3. 

Hence,  by  division,  we  obtain 

2x*+  x3-  20a;2-  7a;  +  24  =  (a;2+  2a;  -  3)  (2a;2-  3x-8), 
and 

2a;4+3ar*-  13a;2-  7a;  +  15  =  (a;2+  2a;  -  3)  (2a;2-    x-o). 

Therefore 
the  L.  C.  M.  =  (a;2  +  2a;  -  3)  (2a;2  -  3a;  -  8)  (2a;2  -  x  -  5). 

We  may  now  prove  the  rule  for  finding  the  least  common 
multiple  of  any  two  compound  Algebraic  expressions. 

Let  A  and  B  denote  the  two  expressions,  F  their  greatest 
common  divisor,  and  M  their  least  common  multiple.  Sup- 
pose that  a  and  b  are  the  respective  quotients  when  A  and 
B  are  divided  by  F;  then 

A  =  aF,     and     B  =  bF.     .     .     .     .   (1) 

Since  F  contains  all  the  factors  common  to  A  and  J3,  the 
quotients  a  and  b  have  no  common  factor,  and  therefore 
their   least   common   multiple    is    «6,    and   hence    the    least 


118      LEAST   COMMON  MULTIPLE   OF  POLYNOMIALS. 

common  multiple  of  aF  and  bF,  or  of  A  and  B,  from  (1)  is 
abF,  by  inspection.     That  is 

M  =  abF (2) 

But  from  (1)  we  have 

AB  =  abFF 

=  MF,  from  (2) (3) 

or  M  =  — -. 

F 

Rule. 

The  least  common  multiple  of  hvo  expressions  may  be  found 
by  dividing  their  product  by  their  G.  O.  D. ,  or,  by  dividing 
either  of  the  expressions  by  their  67.  C.  D.  and  multiplying  the 
quotient  by  the  other. 

From  (3)  we  see  that  the  product  of  any  two  expressions 
is  equal  to  the  product  of  their  G.  C.  D.  and  L.  C.  M. 

To  find  the  least  common  multiple  of  three  expressions,  A, 
J5,  C.  First  find  M,  the  L.  C.  M.  of  A  and  B.  Next  find 
iV,  the  L.  C.  M.  of  M  and  0;  then  N  will  be  the  required 
L.  C.  M.  of  A,  B,  C. 

For  N  is  the  expression  of  least  degree  which  is  divisible 
by  M  and  C,  and  M  is  the  expression  of  least  degree  which 
is  divisible  by  A  and  B.  Therefore  N  is  the  expression  of 
least  degree  which  is  divisible  by  all  three. 

In  a  similar  maimer  we  may  find  the  L.  C.  M.  of  four 
expressions. 

Note.  —  The  theories  of  the  greatest  common  divisor  and  of  the 
least  common  multiple  are  not  necessary  for  the  subsequent  chapters 
of  the  present  work,  and  any  difficulties  which  the  student  may  find 
in  them  may  be  postponed  till  lie  has  read  the  Theory  of  Equations. 
The  examples  however  attached  to  this  chapter  should  be  carefully 
worked,  on  account  of  the  exercise  which  they  afford  in  all  the 
fundamental  processes  of  Algebra. 


EXAMPLES.  119 


EXAMPLES. 

Resolve  into  factors 


1. 

s2  +  xy. 

Ans.  x(x  4-  y) 

2. 

x3  —  x2y. 

x2(x  —  y) 

3. 

10s3  -  25xAy. 

5s3  (2  —  5s?/) 

4. 

x3  —  x2y  +  ^2/2- 

x{x2  —  xy  4-  y2) 

5. 

3a*  _  3as&  +  e«262. 

3a2(a2  -  ab  4-  26s) 

6. 

38a  V  +  57a4s2. 

19a3s2(2s3  4-  3a) 

7. 

ox  —  6s  —  az  4-  62. 

(a  —  b)  (s  —  2) 

8. 

2as  4-  a?/  +  26a;  4-  6?/. 

(2s  4-  y)  (a  4-  6) 

9. 

6s2  4-  3s?/  —  2asc  —  ay. 

(2s  4-  y)  (3s  -  a) 

10. 

2aj*  —  aj8  4-  Ax  —  2. 

(2s  -  l)(s3  +  2) 

11. 

3aB"  4-  5s2  4-  3s  +  5. 

(3s  4-  5)(s2  4-  1) 

12. 

aj*  4.  £3  _j_  2a-  4-  2. 

(s  4-  l)(s3  +  2) 

13. 

y3  -  y2  +  y  -  i- 

(^/-  l)0/2  +  1) 

14. 

20s2  4-  3as?/  -  26a$  -  36?/2 

(2s  4-  3?/)  (as  —  o?/) 

15. 

tT2    _    19x    +    g4i 

(s  -  12)  (s  -  7) 

16. 

a?2  -  19s  4-  78. 

(s  -  13)  (s  -  6) 

17. 

a2  -  Uab  4-  4962. 

(a  —  76)  (a  —  76) 

18. 

a2  4-  oab  4-  662. 

(a  4-  36)  (a  4-  26) 

19. 

m2  -  13m?i  4-  40?i2. 

(m  —  8n)  (m  —  5?i) 

20. 

m2  -  22mn  +  105?i2. 

(m  —  15?i)  (m  —  7?i) 

21. 

s2  -  23s?/  4-  132?/2. 

(s-  12?/)  (s-  11?/) 

22. 

130  4-  31s?/  4-  s2?/2. 

(26  4-  s?/)(5  4-  xy) 

23. 

132  -  23s  4-  s2. 

(12  -  s)(ll  -  s) 

24. 

88  4-  19s  4-  s2. 

(8  4-  s)(ll  4-  s) 

25. 

65  4-  8s?/  —  x2y2. 

(5  4-s?/)(13  -  s?/) 

26. 

x*  +  16s  -  260. 

(s  4-  26)  (s  -  10) 

27. 

x*  _  Haj  _  26. 

(s  4-  2)(s  -  13) 

28. 

a2b2  -  3o6c  -  10c2. 

(a6  4-  2c)  (a&  -  5c) 

29. 

^  _  aV  -  132a4. 

(s2  4-  11a2)  (s2  -  12a2) 

30. 

4s2  4-  23s  4-  15. 

(s  4-  5)  (4s  4-  3) 

31. 

12s2  -  23s.y  4-  10?/2. 

(3s  -  2y)  (4s  -  by) 

32. 

8s2  -  38s  4-  35. 

(2s  -  7)  (4s  -  5) 

33. 

12s2  -  31s  -  15. 

(12s  4-  5)(s  -  3) 

120  EXAMPLES. 

34.  3  +  11a;  -  4a;2.  Ans.  (3  -  x)(l  +  4a;). 

35.  6  +  5x  -  6a;2.  (2  +  3.x)  (3  -  2a;). 

36.  4  -  5a;  -  6x2.  (4  +  3a;)  (1  -  2a;). 

37.  5  +  32a;  -  21a;2.  (1  +  7a;)  (5  -  3a;). 

38.  20  -  9a;  -  20a;2.  (5  +  4a;)  (4  -  ox). 

39.  (1811)2  -  (689)2.  2500  x  1122  =  2805000. 

40.  (8133)2  -  (8131)2.  16264  x  2  =  32528. 

41.  (24a;  +  yY  -  (23a;  -  y)2.  47a; (.-c  +  2//). 

42.  (5a;  +  2y)2  -  (3.x  -  y)2.  (8a;  +  y)(2x  +  3y). 

43.  9a;2  -  (3a;  -  by)2.  5y(6as  -  by). 

44.  16a;2  -  (3a;  +  l)2.  (7a;  +  l)(x  -  1). 

45.  a6  +  72963.  (a2  +  96)  (a4  -  9a*b  +  8162). 

46.  xhf  -  512.  (xy  -  8)(x2y2  -f  Sxy  +  64). 

47.  500a;2?/  -  20ys.  20y(bx  +  y)(bx  -  y). 

48.  (a+by-1.  [(a+6)2+l](«+6+l)(a  +  6-l). 

Find  the  greatest  common  divisor  of 

49.  GGaW,  44aW,  24aW.  2aW. 

50.  x2  +  x,  (x  +  l)2,  a;3  +  1.  a;  +  1. 

51.  a;8  -f  8?/3,  a;2  +  xy  —  2y2.  x  -f-  2?/. 

52.  12a;2  +  x  -  1,  15a;2  +  8a;  +  1.  3a;  +  1. 

53.  2a;2  +  9a;  +  4,  2a;2  +  Hx  +  5,  2a;2-3a;-2.  2»+l. 

54.  3a;4-3a;3-2a;2-a;-l,  6a;4-3a;3-a;2-a;-l.  3a;2+l. 

55.  2a;3  -  9ax*  +  9a2a;  -  7a3,  4a;3  -  20aa;2  +  20a2a;  -  16a3. 

Ans.  x2  —  ax  -f-  a2. 

56.  4a;5-fl4a;4+20a;3+  70a;2,  8a;7  +  28a;6-  8a;5-  12s4 +  56a8. 

Ans.  2ar  (2a;  4-7) 
Find  the  least  common  multiple  of 

57.  35aa;2,  42a?/2,  30az2.  210ax2y2z2. 

58.  a;2  -  3a;  +  2,  a;2  -  1.  (x  +  l)(.r  -  1 )  (a;  -  2).. 

59.  a?a  -  5a;  +  4,  a;2  -  6a  +  8.  (a;  —  4)  (a?  —  1)  (a;  —  2) . 

60.  a;2  -  1,  a;3  +  1,  a;3  -  1.  xc'  -  1. 

61.  a;2  -  1,  ar  +  1,  a;4  +  1,  a;8  -  1.  a;8  -  1. 

62.  X2  -   1,  SC8  +   1,  K8  -   1,  X*  +   1.  .r1'-  -    1. 

63.  »8+2«2-3aJ,  2a;8  +  5ar-3a;.      »(»-!) (aj+3)(2aj-l). 


A    FRACTION — ENTIRE   AND   MIXED    QUANTITIES.    121 


CHAPTER    VIII. 

FRACTIONS. 

Note. — In  this  chapter  the  student  will  find  that  the  definitions, 
rules,  and  demonstrations  closely  resemble  those  with  which  he  is 
already  familiar  in  Arithmetic. 

78.   A  Fraction  —  Entire  and  Mixed  Quantities. 

—  A  Fraction  is  an  expression  of  an  indicated  quotient  by 
writing  the  divisor  under  the  dividend  with  a  horizontal  line 
between  them  (Art.  11).  In  the  operation  of  division  the 
divisor  sometimes  may  be  greater  than  the  dividend,  or  may 
not  be  contained  in  it  an  exact  number  of  times  ;  in  either 
case    the    quotient    is    expressed    by  means    of   a   fraction. 

Thus,  the  expression  -  indicates  either  that  some  unit  is 
b 

divided  into  b  equal  parts,  and  that  a  of  these  are  taken,  or 
that  a  times  the  same  unit  is  divided  into  b  equal  parts,  and 
one  of  them  taken. 

In  an}'  fraction  the  upper  number,  or  the  dividend,  is 
called    the  numerator,   and    the   lower  number,   or  divisor, 

is  called  the  denominator.     Thus  in  the  above  fraction  - 

b' 
which  is  read  a  divided  by  6,  a  is  called  the  numerator  and 
b  the  denominator,  and  the  two  taken  together  are  called  the 
terms  of  the  fraction.  Thus  the  denominator  indicates  into 
how  many  equal  parts  the  unit  is  to  be  divided,  and  the 
numerator  indicates  how  many  of  those  parts  are  to  be 
taken. 

Every  integer  or  integral  expression  may  be  considered  as 
a  fraction  whose  denominator  is  unit}' ;  thus, 

a  ,    7         a  4-  b 

a  =  -,     a  -f-  b  =  — ?— . 


122     TO   REDUCE   A    FRACTION    TO   ITS    LOWEST   TERMS. 

An  entire  quantity  or  integral  quantity  is  one  which  has  no 

fractional  part ;  as  ab  or  a2  —  2ab. 

A  mixed  quantity  is  one  made  up  of  an  integer  and  a 

x 

fraction  ;  as  b  -\ — . 

a 

A  proper  fraction  is  one  whose  numerator  is  less  than  its 
denominator ;  as 


a  +  x 
An  improper  fraction  is  one  whose  numerator  is  equal  to 

or  greater  than  the  denominator  ;  as  -  and  -       x. 

a  a 

The  reciprocal  of  a  fraction  is  another  fraction  having  its 
numerator  and  denominator  respectively  equal  to  the  denom- 
inator and  numerator  of  the  former. 

79.  To  Reduce  a  Fraction  to  its  Lowest  Terms. 

—  Let  -  denote  any  fraction,  and  — -  denote  the  same  frac- 
b  mb 

tion  with  its  terms  multiplied  by  m. 

Now  -  means  that  a  unit  is  divided  into  b  equal  parts,  and 

that  a  of  these  are  taken (1) 

And  —  means  that  the  same  unit  is  divided  into  mb  equal 
mb 

parts,  and  that  ma  of  these  are  taken (2) 

Hence  b  parts  in  (1)  =  mb  parts  in  (2). 

.*.     1  part    in  (1)  =  m    parts  in  (2), 

and  .*.     a  parts  in  (1)  =  am  parts  in  (2), 

that  is, 
Conversely, 

Therefore,  the  value  of  a  fraction  is  not  altered  if  the 
numerator  and  denominator  are  either  both  multiplied  or  both 
divided  by  the  same  quantity. 


a 
b 

= 

ma 
mb 

ma 
mb 

= 

a 
b' 

EXAMPLES.  123 

When  both  numerator  and  denominator  are  divided  by 
all  the  factors  common  to  them,  the  fraction  is  said  to  be 
reduced  to  its  loicest  terms. 

Hence  to  reduce  a  fraction  to  its  lowest  terms  we  have  the 
following 

Rule. 

Divide  both  numerator  and  denominator  by  their  greatest 
common  divisor. 

Dividing  both  terms  of  a  fraction  by  a  common  factor  is 
called  canceling  that  factor. 

EXAMPLES. 

Reduce  the  following  fractions  to  their  lowest  terms. 
6a2bc2       2ac 


b 


cJab'2c         3b 

The  greatest  common  divisor  3abc  of  both  terms  is  can- 
celed. 

g-^    lx2yz    =  J_ 
\J  2Sx3yz2       Axz 

The  factor  Ixhjz,  which  is  the  greatest  common  divisor  of 
both  terms,  is  canceled. 

24a3c2x2  24asc2x2  4ac2 


180*3?  -  12oV       Sa2x2(3a  -  2x)       3a  -  2x 
Here  6a2x2  is  canceled  since  it  is  the  greatest  common 
divisor  of  both  terms. 

Note.  —  In  each  of  these  examples,  the  resulting  fractions  have 
the  same  value  as  the  given  fractions,  but  they  are  expressed  in  a 
simpler  form.  The  student  should  be  careful  not  to  begin  canceling 
until  he  has  expressed  both  terms  of  the  fraction  in  the  most  con- 
venient form,  by  factoring  when  necessary.     Thus, 

[Q  ex2  -  Sxy  -=  2x(Sx  ~  *a)  =  ^ 

9xy  —  12y2       3y(3x  -  Ay)        3y 
Instead   of   reducing  a  fraction   to   its   lowest   terms  by 
dividing  the  numerator  and  denominator  by  their  G.  C.  D., 


124  EXAMPLES. 

we  ma}-  divide  by  any  common  factor,  and  repeat  the  process 
till  the  fraction  is  reduced  to  its  lowest  terms.     Thus, 

:     24a  W       12a2bc2       6ac       2a 


36aW       18ab2c2       9bc       36 
ce  the 

Scrbc2 
12ab2cd 
3a2  -  Gab 


Reduce  the  following  fractions  to  their  lowest  terms. 


2crb  -  4ce6a 

2 


~      Ax2  —  Oy 
4x2  -f  6xy 
£        20  (x3  -  y3) 


10. 


5a2  +  5xy  +  by2 
x3  —  2xy2 


Ans.  — — . 
Sbd 

3 

2b 

2x 

-Sy 

2x 

4(a 

-y)- 

X 

X2 

-  2?y2' 

x*  —  4a2?/2  -J-  4?/4 

When   the   factors    of    the    numerator   and   denominator 

cannot    be    found    by    inspection,    their    greatest    common 

divisor  may  be  found  by  the  rule  (Art.  74) ,  and  the  fraction 

then  reduced  to  its  lowest  terms. 

„     ~   ,         .     ..    ,         .,  3a3  -  13a2  +  23a  -  21 

11.    Reduce  to  its  lowest  terms  -" . 

15a3  -  38a2  -  2x  +  21 

The  G.  C.  D.  of  the  numerator  and  denominator  is  3a  —  7. 
Dividing  the  numerator  and  denominator  by  3a  —  7,  we 
obtain  the  respective  quotients  a2  —  2a  +  3  and  5a2  —  a  —  3. 
Therefore 

3a3-13a2+23a-21  =  (3a  -7)  (a2-  2a  -V3)  _  a2- 2a  4- 3 
15a3- 38a2- 2a +  21  ~  (3a  -7)  (5a2-  a  -  3)  "  5a2- a -3' 

This  example  may  also  be  solved  without  finding  the 
G.  C.  D.  by  the  rule  (Art.  74)  as  follows : 

By  2  of  Art.  74,  the  G.  C.  D.  of  the  numerator  and 
denominator  must  divide  their  sum  18s8  —  51a2  -f-  21a,  that 
is,  3a (3a  —  7)  (2a  —  1).  If  there  be  a  common  divisor  it 
must  clearly  be  3a  —  7.     Hence,  arranging  the  numerator 


TO   REDUCE   A    MIXED    QUANTITY    TO   A    FRACTION.    1 25 

and  denominator  so  as  to  show  8a  —  7  as  a  factor,  we  have 
the  fraction 

_  a?(3x  -  7)  -  2x(3x  -  7)  +  3  (3a;  -  7) 

5a?  (3a;  —  7)  —  a?(3a;  —  7)  —  3  (3a;  —  7) 

=  (3a;  -  7)  (a;2  -  2x  +  3)  =  x2  -  2a;  +  3 

(ox  —  7)  (oar  —  x  —  3)        bx*  —  x  —  3 

When  either  the  numerator  or  denominator  can  readily  be 

factored  we  may  use  the  following  method  : 

12.  Reduce  to  its  lowest  terms  ■ —        — '/    ~ — ■ . 

7a;3  -  18ar  +  6x  +  5 

The  numerator  =  x(x2  +  3a;  -  4)  =  x(x  +  4)(x  -  1). 

The  only  one  of  these  factors  which  can  be  a  common 
divisor  is  x  —  1,  since  the  denominator  does  not  contain  a;, 
and  5  the  last  term  in  the  denominator  does  not  contain  4. 
(See  Art.  66.)  Hence,  arranging  the  denominator  so  as  to 
show  x  —  1  as  a  factor, 

the  fraction  =  — *(*+*j(*-l) =     *(*+*)    . 

7a;2(a;-l)-lla-(aj-l)-5fa-.-l)       7. 

Reduce  to  lowest  terms 

13.    : .  Ans.  . 

a  4-  26 

14-  ^ 

15. 


a3  —  a2b  —  ab2 

2b3 

a3  +  3d2b  +  3ab2 
x3  —  5a;2  +  7x  — 

+ 
3 

2b3' 

x3  -  2>x  4-  2 
Aa3  4-  I2a*b  -  ah2 

-  15b3 

x  4-  2 

2a  4-  5b 

6a3  4    13a26  -  Aab2  -  15b3'  3a  4-  5b 


80.   To  Reduce  a  Mixed  Quantity  to  the  Form 

be 

of  a  Fraction.  —  Let  it  be  required  to  reduce  a  H 

b  4-  c 

to  the  form  of  a  fraction. 

The  entire  part  a  =  -  (Art.  78)  =  a(h  +  c)  (Art.  79). 

1  6  +  c 

Heuce  a  +  _»2_  =  "<ft  +  c)  +  -»£_ 

6  4-  c  6  4-  c  6  4-  c 

_  ab  4-  ftc  4-  6c 
64c 


126    TO   REDUCE   A    ERACTION   TO   A    illXED   QUANTITY. 

Hence  we  have  the  following 

Rule. 

Multiply  the  entire  part  by  the  denominator,  and  to  the 
product  add  the  numerator  with  its  proper  sign;  under  this 
sum  place  the  denominator,  and  the  result  will  be  the  fraction 
required. 


EXAMPLES. 

Reduce  the  following  to  fractional  forms : 

1                    ,        & 
1.    a  —  x  -\ . 

a  +  x 

Ans.  — , 

a  -f-  x 

2-"+1+,-3- 

x2  -  2x  -f  1 

x  —  3 

Xs 

o.    Jb       ■    xy    -\-  y      ■ 

X  +  y 

X  +   ?/' 

A                  ,7,             «2    +    O2 

4.    a  +  b ! . 

2b2 

81.  To  Reduce  a  Fraction  to  an  Entire  or  Mixed 

(ix  -|-   2 1*2 
Quantity.  —  Let  it  be  required  to  reduce  - — — — —  to  a 

a  +  x 

mixed  quantity. 

Performing  the  division  indicated,  we  have 
ax  -f  2x2  _  x2 

—  x   -f- 


a  -f  x  a  -{-  x 

Hence  we  have  the  following 

Rule. 

Divide  the  numerator  by  the  denominator,  as  far  as  possible, 
for  the  entire  part,  and  annex  to  the  quotient  a  fraction  having 
the  remainder  for  numerator ,  and.  the  divisor  for  denominator ; 
it  will  be  the  mixed  quantity  required. 


TO   REDUCE   FRACTIONS    TO   THEIR  L.    C    D.  127 


EXAMPLES. 

Reduce  to  whole  or  mixed  quantities  the  following : 

1.    — .  Aits,  da  H 

7  7 

2     a8  +  3ab  a  +     2o6 


a  +  b  a  +  b 

360C  +  4Q  4f(c       4c 

9  9 

4.  8«1±3&  2a  +  36 

4  a  4a 

5.  *  +  »+».  x  + 


35  +  3  a  +  3 

G.    2*2  -6^-1.  2c  -       X 


JB  —  3  a;  —  3 

^    +    ^    +    £+1+  2 


a;  —  1  95—1 

82.  To  Reduce  Fractions  to  their  Least  Common 

Denominator.  —  Let  it  be  required  to  reduce  — ,  — ,  —  to 

yz   zx   xy 

equivalent  fractions  having  the  least  common  denominator. 

The  least  common  multiple  of  the  denominators  is  xyz. 
Dividing  this  L.  C.  M.  by  the  denominators,  yz,  zx,  and  xy, 
we  have  the  quotients  x,  y,  and  z,  respectively. 

By  Art.  79,  both  terms  of  a  fraction  may  be  multiplied  by 
the  same  number  without  altering  its  value  ;  therefore  we 
may  multiply  both  terms  of  the  first  fraction  by  x,  both  terms 
of  the  second  fraction  by  y,  and  both  terms  of  the  third 
fraction  by  z,  and  the  resulting  fractions  will  be  equivalent 
to  the  given  ones. 

tt  a_  __  cm       b_  __   by_       c_  _    cz_ 

yz       xyz*     zx       xyz1     xy       xyz 

(ix      bi/  cz 

That  is,  the  resulting  fractions  — ,  —%-,    and  —  have  the 

xyz    xyz  xyz 

same  values  respectively  as  the  given  fractions  — ,  — ,  and  — , 

yz  zx  xy 

and  they  have  the  least  common  denominator  xyz.     Hence, 


128  RULE   OF  SIGNS    IN   FRACTIONS. 

for  reducing    fractions  to  their  least  common  denominator, 
we  have  the  following 

Rule. 

Find  the  least  common  multiple  of  the  given  denominators, 
and  take  it  for  the  common  denominator ;  divide  it  by  the 
denominator  of  the  first  fraction,  and  multiply  the  numerator 
of  this  fraction  by  the  quotient  so  obtained;  and  do  the  same 
with  all  the  other  given  fractions. 

Note  1.  —  It  is  not  absolutely  necessary  to  take  the  least  common 
denominator.  Any  common  denominator  may  be  used.  But  in 
practice  it  will  be  found  advisable  to  use  the  least  common  denomina- 
tor, as  the  work  will  thereby  be  shortened. 

Note  2.  —  It  frequently  happens  that  the  denominators  of  the 
fractions  to  be  reduced  do  not  contain  a  common  factor.     Thus,  the 

?   I  / 

therefore  the  least  common  denominator  of  these  fractions  is  bdf,  the 
product  of  all  their  denominators. 

EXAMPLES. 

Reduce  the  following  fractions  to  their  L.  C.  D. 

Ans    —     —     — 
US'  V2x3'    12z3'    12a,*3' 

a(x  -\-  a)    x(x  +  a)         a2 

x'2  —  a2        x2  —  a2      x2  —  a2 

83.  Rule  of  Signs  in  Fractions.  —  The  signs  of  the 
several  terms  of  the  numerator  and  denominator  of  a  fraction 
relate  only  to  those  terms  to  which  they  are  prefixed,  while 
the  sign  prefixed  to  the  dividing  line  relates  to  the  fraction 
as  a  whole,  and  is  the  sign  of  the  fraction.     Tims,   in  the 

fraction  — — - — ,  the  sign  of  a,  the  first  term  of  the  numer- 
a  +  b 

ator,  is  -f  understood,  the  sign  of  the  second  term  b  is  — , 

and  the  sign  of  each  term  "  and  b  of  the  denominator  is  +> 

while  the  sign  of  the  fraction  itself  is  — . 


1. 

2. 

3       4         5 
4ic'    Go;2'    12z3' 

a           x            a2 

x  —  a    x  —  a    x2  —  a2 

RULE    OF  SIGNS    IN    FRACTIONS.  129 


The  symbol  means  the  quotient  resulting    from    the 

—  b 

division  of  —a  by  —  b\  and  this  is  obtained  by  dividing  a 

by  b,  and  prefixing  -{- ,  by  the  rule  of  signs  in  division  (Art. 
40). 

Therefore  —  =  +-  =  - (1) 

Also,  ^—  is  the  quotient  of  —  a  divided  by  b  ;  and  this  is 

obtained  by  dividing  a  by  b,  and  prefixing  — ,  by  the  rule  of 

signs. 

Therefore  ^—  =  — - (2) 

In  like  manner,  — —  is  the  quotient  of  a  divided  by  —  b ; 
—  b 

and  this  is  obtained  by  dividing  a  by  b,  and  prefixing  — ,  by 

the  rule  of  signs. 

Therefore  —  =  -- (3) 

—  b  b 

Hence,  we  have  the  following  rule  of  signs  : 

(1)  If  the  signs  of  both  numerator  and  denominator  be 
changed,  the  sign  of  the  whole  fraction  remains  unchanged. 

(2)  If  the  sign  of  the  numerator  alone  be  changed,  the  sign 
of  the  whole  fraction  will  be  changed. 

(3)  If  the  sign  of  the  denominator  alone  be  changed,  the 
sign  of  the  whole  fraction  will  be  changed. 

Or  they  may  be  stated  as  follows : 

(1)  We  may  change  the  sign  of  every  term  in  the  numera- 
tor and  denominator  of  a  fraction  without  altering  the  value 
of  the  fraction. 

(2)  We  may  change  the  sign  of  a  fraction  by  changing  the 
sign  of  every  term  either  in  the  numerator  or  denominator. 


130  EXAMPLES. 

EXAMPLES. 

a  —  b         —a  -f-  b 


1. 


3. 


—  m  -f-  » 
—  b  +  ft 
2a 
3a? 


b 

— 

a 

n 

— 

m 

a 

-  b 

2x 

Sx 

-  xz  —x  -|-  xL  x*  —  x 

The  intermediate  steps  may  usually  be  omitted. 
From  Art.  36  we  have 
abmx  _  (  —  a)  (  —  b)(—  m)x  _  a(  —  b)(—m)x  _    . 
pqr  (-P)qr  (-p)(—V)r 

That  is,  if  the  terms  of  a  fraction  are  composed  of  any 
number  of  factors,  any  even  number  of  factors  may  have 
their  signs  changed  without  altering  the  value  of  the  fraction; 
but  if  any  odd  number  of  factors  have  their  signs  changed, 
the  sign  of  the  fraction  is  changed. 

Thus,    (a  -  6)  (b  -  c)  _  (b  -  a)  (b  -  c)  =  (b  -  a)  (c  -  6)  > 

(x-y)(y-%)    O/-a0(2/-2)    (y-x)(z-y)' 

When  the  numerator  is  a  product,  any  one  or  more  of  its 
factors  can  be  removed  from  the  numerator  and  made  the 
multiplier. 

rp,  abed  i    cd  7    ,     1 

Thus,  =  ab =  abed- . 

a  +  b  a  +  b  a  -f  b 

Change  the  signs  of  the  following  fractions  so  as  to 
express  them  altogether  in  four  different  ways. 

,     ft  —  b  .       b  —  ft       b  —  a       a  —  h 

4.    .  Ans. , , . 

x  —  y  y  —  xx  —  yy  —  x 

E  X  —X  —X  X 


y  -  z  z  -  if      y  -  z       z  -  y 

6  —  ft  a  —  b  a  —  b  b  —  a 


7. 


a  —  b  -\-  c  b  —  c  —  a       a  —  b  -\-  c       b  —  c  —  ft 

abed  a(  —  h)(  —  c)d,    a(  —  b)rd  abed 

xyz'  xyz  '   xy(—  z)  '  x(—  y)  (— z) 


ADDITION   AND   SUBTRACTION    OF  FRACTIONS.        131 

84.  Addition  and  Subtraction  of  Fractions.  —  Let 

it  be  required  to  add  together  -  and  -. 

c  c 

Here  the  unit  is  divided  into  c  equal  parts,  and  we  first 

take  a  of  these  parts,  and  then  b  of  them;    i.e.,  we  take 

a  4-  b  of  the  c  parts  of  the  unit ;  and  this  is  expressed  by 

the  fraction  — — — . 


a 
c 

+ 

b 

c 

= 

a 

+  b 

c 

a 

b 



a 

-  b 

c 

c 

c 

Similarly 

a  c 

Let  it  be  required  to  add  together  -  and  -. 

0  (X 

-rrr      ,  a  ttd  ,     C  &C 

vv  e  have  -  =  — ,  and  -  =  — . 

b       bcl  d       bd 

Here  in  each  case  we  divide  the  unit  into  bd  equal  parts, 
and  we  first  take  ad  of  these  parts,  and  then  be  of  them ; 
i.e.,  we  take  ad  +  be  of  the  bd  parts  of  the  unit;  and  this 

is  expressed  by  the  fraction 


bd 

a 
b 

+ 

c 
d 

ad  -h  be 
bd 

a 

b 

— 

c 
d 

ad  —  be 
bd 

Similarly 

Here  the  fractions  have  been  reduced  to  a  common  denom- 
inator bd.  But  if  b  and  d  have  a  common  factor,  the  product 
bd  is  not  the  least  common  denominator,  and  the  fraction 

- — =t— -  will  not  be  in  its  lowest  terms.     To  avoid  working 
bd 

with  fractions  which  are  not  in  their  lowest  terms,  it  will  be 

found  advisable  to  take  the  least  common  denominator,  which 

is  the  least  common  multiple   of   the  denominators  of   the 

given  fractions  (Art.  82,  Note  1). 


132  EXAMPLES. 

Hence  we  have  the  following 
Rule. 

To  add  or  subtract  fractions,  reduce  them  to  the  least 
common  denominator ;  add  or  subtract  the  numerators,  and 
write  the  result  over  the  least  common  denominator. 

EXAMPLES. 

i      X  -it  a  -\-  c   a  —  c        ,  a  -+-  d 

1 .    Arid  — :z— ,  ,  and  — ; — . 

b  b  b 

Here  the  fractions  have  already  a  common  denominator, 
and  therefore  need  no  reducing.     Hence  we  have 

a  +  c       f  —  c      a  -\-  d  _  a  +  c  +  a  —  c  +  a  +  d  _  3a  +  d 
b  b  6     "  b  "~      b 

-2.    Add  2-^±^  and  *JLz^. 
3a  9a 

Here  the  least  common  denominator  is  da.  Hence  we 
have  (Art.  82) 

2x  +  a    .    5x  —  4a  _  3 (2x  +  a)    ,    5a;  —  4a 
3a  9a  9a  9a 

6rc  +  3a  +  5x  —  4a        11a;  —  a 


9a  9a 


0     ^        4a  —  26  .  ,     3a  —  3b 
3.    From  -  -  take  - 


c  c 

4a  -  2b  _  3a  -  3b  _  4a  -  26  -  (3a  -  3b) 
c  c  c 

4a  —  26  —  3a  +  36       a  +  6 


Note  1.  —  To  insure  accuracy,  the  beginner  is  recommended  to  put 
down  the  work  in  full;  and  when  a  fraction  whose  numerator  is  not  a 
monomial  is  preceded  by  a  —  sign,  he  is  recommended  to  enclose  Its 
numerator  in  a  parenthesis  us  above  before  combining  it  with  the 
other  numerators. 


EXAMPLES.  133 


4:.)  Add  — ^—  and 


a  ,      b 


b  —  a 


Here  — —  = —  (Art.  83), 

b  —  a  a  —  b 

a  b  a  b 


, 


a  —  b       b  —  a       a  —  b       a 

_  a  —  b  _  . 

"  a  -  b  ~ 

Add  *~*a  3y  ~  q,  and  2a  ~  3* 


The  least  common  denominator  is  «a^. 

#  —  2y       Sy  —  a       2a  —  3x 
xy  ay  ax 

a(x  -  2y)  +  «(3^  -  a)  +  y(2a  -  3x) 


axy 
ax  —  2ay  -f-  3a?/  —  ax  -\-  2ay  —  3xy 


=  0, 


axy 
since  the  terms  in  the  numerator  destroy  each  other. 


-g?  From  ^+-^  take  * 


a  —  b  a  -\-  b 

The  least  common  denominator  is  a2  —  b2. 

•      a  +  6  _  a  ~  b  --   (a  +  &)2  _  (g  ~  6)2 
a  -  6        a  +  6         a2  -  62  a2  -  62 

_  a2  +  2ab  +  b2  -  (a2  -  2ab  +  b2) 
a2  -  b2 

=      4ab 
a2  -  br 


^   Add  ?^^  and 


x  —  2a  x  —  a 

The  least  common  denominator  is  (x  —  2a)  (x  —  a). 


134  EXAMPLES. 

Hence  we  have 
2x  -  3a  _  2x  —  a  _  (2s  —  3a)  (x  —  a)  —  (2x  —  a)  (x  —  2a) 
x  —  2a        x  —  a  (x  —  2a)  (x  —  a) 

_  2x2  -  bax  +  3a2  -  (2x2  -  bax  +  2a2) 

(x  —  2a)  {x  —  a) 
_  2z2  -  bax  +  3a2  -  2x2  +  5a.x  -  2a2 
(x  —  2a)  (x  —  a) 
a2 
(a:  —  2a)  (a;  —  «)' 

Note  2.  —  In  finding  the  value  of  an  expression  like 
— (2x  —  a){x  —  a), 
the  beginner  should  first  express  the  product  in  parentheses,  and  then 
after  multiplication,  remove  the  parentheses,  as  we  have  done.     After 
a  little  practice  he  will  be  able  to  take  both  steps  together. 

Note  3.  —  In  practice,  the  foregoing  general  method  may  some- 
times be  modified  with  advantage.  When  the  sum  of  several  fractions 
is  to  be  found,  it  is  often  best,  instead  of  reducing  at  onoe  all  the 
fractions  to  their  L.  C.  D.,  to  take  two  or  more  of  them  together,  and 
combine  the  results. 

o     c-      i.-    a  +  3       a  4-  4  8 

8.    Simplify  — ! ! — . 

1      J  a  -  4       a  -  3        a2  -  16 

Here,  instead  of  reducing  all  the  fractions  to  the  least 

common  denominator  at  once,   we  may  take  the   first  two 

fractions  together,  as  follows  : 

a  +  3  _  q  +  4  _        8        _  q2-9-(a2-16) 

a  — 4       a  — 3       a2— 10 


9.    Simplify 


(a  —  4)  (a  —  3)          a2— 16 

7                               8 

(a  — 4)  (a  — 3)      (a  +  4)(a- 

-J) 

o2  —  H 

(a  -  4)  (a  +  4)  (a  -  3) 

1                 2x                4x* 

a  —  x       a  -f-  x       a~  +  %*       "4  +  #4 

Here  we  see  that  the  first  two  denominators  give  L.  C.  D. 
a2  —  x2  ;  and  this  with  <r  +  X*  gives  L.  ('.  1).  a4  —  x4 ;  and 
this  with  a4  -f-  x*  gives  L.  C.  I).  ab  —  xH.      Hence,  instead 


EXAMPLES.  135 

of  reducing  all  the  fractions  to  the  least  common  denom- 
inator, we  proceed  as  follows  : 


The  first  two  terms 
The  first  three  terms 


a  +  x  —  (a  —  x)  _         2x 


a2 

-  X2 

a- 

-   X2' 

2x 

2a- 

a2  - 

X2 

a2 

+  x2 

2x(cr 

+ 

X2)- 

-  2x(a2 

— 

X2) 

a4  ■ 

-  a4 

Ax* 


The  whole  expression 


a" 

-  z4 

4x? 

a* 

-   X* 

8x7 

Ax3 


a4  +  x* 


Xs 


10.   Simplify -+ + . 

1     J  (a-b)(a-c)      (b-c)(b-a)     (c— a)(c— 6) 

The  beginner  is  very  liable,  in  this  example,  to  take  the 
product  of  the  denominators  for  the  least  common  denom- 
inator, and  thus  to  render  the  operations  very  laborious. 
The  denominator  of  the  second  fraction  contains  the  factor 
b  —  a,  and  this  factor  differs  from  the  factor  a  —  6,  which 
occurs  in  the  denominator  of  the  first  fraction,  only  in  the 
sign  of  each  term.  Also  the  denominator  of  the  third 
fraction  contains  the  factors  c  —  a  and  c  —  &,  which  differ 
from  the  factors  a  —  c  and  b  —  c  in  the  denominators  of  the 
first  two  fractions,  only  in  the  signs  of  each  term.  It  is 
better  to  arrange  these  factors  so  that  a  precedes  b  or  c,  and 
that  b  precedes  c.     By  Art.  83  we  have 

b  b 


and 


(b  -  c)(b  -  a)  "        (b  -  c)(a  -  &)' 
c  c 


(c  —  a)(c  —  b)        (a  —  c)(b  —  c) 
Hence  the  given  expression  may  be  put  in  the  form 
a  b  c 


(a  -  b)(a  -  c)         (b  -  c)(a  -  b)         (a  -  c)(b  -  c) 
whose  L.  C.  D.  we  see  at  once  is  (a  —  b)  (a  —  c)  (6  —  c). 


136  EXAMPLES. 

Reducing  the  fractions  to  the  least  common  denominator, 
the  given  expression  becomes 

a(b  —  c)  —  b(a  —  c)  -f  c(a  —  b) 
(a  —  b)  (a  —  c)(b  —  c) 

_  ab  —  ac  —  a&  +  be  -f  ac  —  6c  _  n 
(a  -  b)  (a  -  c)  (b  -  c)         "~     " 

The  work  is  often  made  easier  by  completing  the  divisions 
represented  by  the  fractions. 

11.    Simplify  1  +  2fl?  +  1  -  ^±-5. 
J  2a;  -  2       2a;  +  2 

We  have  by  division 

1  +  1  +  — §—    -  2  -  3 


2x  -  2  2a;  +  2        2a  -  2        2a;  +  2 

-  3a;  +  3  -  a;  -f  1 

2or  -  2 

-  g  ±  2 

a;2  -  l' 

12     5a;  ~  1  _  3a;  —  2       ac  —  5  ,__   25a;  —  61 

8  7  4" 

2a;  +  5  _  x  +  3  _  27 
a;  2a;  8a;'2 

x  —  4  _  x  —  7 
x  —  2       a;  —  5 

4a'  +  fr2  _  2a  -  & 
4a2  -  &2       2a  -f-  6* 

^    _J> 3a2_  _  4  -  13s 

1  +  2a;        1  -  2x        1   -  4a;2' 


18. 


a;  —  2        3a;  +  6        a-2  —  4 

3a  2  2 

1  —  a2       a;  —  1        a;  +  1* 


Ql 


+ 


56 

12a;2  +  28a 

-  27 

8a;2 

6 

(a;  -  2)  (x 

-  5) 

lab 

4a2 

-  lr 

1 

-  6a-2 

1 

-  4a;2' 

2(13a;  + 

<) 

3(3  -  2)  (a; 

+  2) 

7a; 

1 

-   X1 

1 

0. 

(a—b)  (<t-c)       (6-e)  (&-«)        (<-•-«)  {c~b) 


TO   MULTIPLY  A    FRACTION   BY  AN   INTEGER.        137 

85.  To  Multiply  a  Fraction  by  an  Integer. — 

Rule. 

Multiply  the   numerator  by  that    integer;    or  divide  the 
denominator  by  that  integer. 

The  rule  may  be  proved  as  follows : 

(1)  Let  -  denote  any  fraction,  and  c  any  integer;    then 

will  -   x  c  —  — ;  for  in  each  of  the  fractions  -  and  —  the 
b  b  b  b 

unit  is  divided  into  b  equal  parts,  and  the  number  of  parts 

taken  in  —  is  c  times  the  number  taken  in  -. 
b  b 

(2)  Let  -—  denote  any  fraction,  and  c  any  integer ;  then 

OG 

_-XC   =   -,by(l), 

=  %  (Art.  79). 
b 

86.  To  Divide  a  Fraction  by  an  Integer. — 

Rule. 

Divide  the  numerator  by   that   integer;    or  multiply   the 
denominator  by  that  integer. 

The  rule  may  be  proved  as  follows : 

ac 

(1)  Let  —  denote  any  fraction,  and  c  any  integer;  then 

will  — -  -T-  c  =  -  ;  for  —  is  c  times  -  (Art.  85)  ;  and  there- 
o  b  b  b 

fore  -  is  the  quotient  of  —  divided  by  c. 
b  b  J 

(2)  Let  -  denote  any  fraction,  and  c  any  integer ;  then 

?  =  f£(Art.79). 

a  ac  «    i     /i\ 

b  be  be 


138  TO   MULTIPLY  FRACTIONS  —  EXAMPLES. 

87.    To  Multiply  Fractions.  —  Let  it  be  required  to 


multiply  ^  by 

c 
d 

T.  .  a 
Put  -  =  £, 
b 

and  -  = 
d 

:  y> 

Then 

(Art.  85; 

a  = 

bx, 

and 

c  =  dy; 

therefore 

ac  = 

bdxy ; 

divide  by  bd, 

ac  _ 
bd  ~ 

xy. 

But 

xy  = 

a       c 
b  X  d* 

.•. 

a 
b 

*i= 

and  ac  is  the  product  of  the  numerators,  and  6rf  the  product 
of  the  denominators.     Hence  the  following 

Rule. 

Midtiply  the  numerators  together  for  the  numerator  of 
the  product,  and  the  denominators  for  the  denominator  of  the 
product. 

Similarly,  the  rule  may  be  demonstrated  when  more  than 
two  fractions  are  multiplied  together. 

Note.  —  If  either  of  the  factors  is  a  mixed  quantity,  it  is  usually 
best  to  reduce  it  to  a  fractional  form  before  applying  the  rule.  Also, 
it  is  advisable  to  indicate  the  multiplication  of  the  numerators  and 
denominators,  and  to  examine  if  they  have  common  factors;  and,  it 
so,  to  cancel  them  before  performing  the  multiplication. 


EXAMPLES. 

1.    Multiply  together  ^  and  — . 
46  9a 

3a        8c  =  §a_X_8c  _  2_c   (Art   79) 
46       Uo       46  x  i)a       36   v  ' 


EXAMPLES.  I3g 

o     -m-  i±-  1         2n2       ,       (a  +  b)2 

2.    Multiply  — by  J rm    '  . 

J!/  '  J  a-  -  b2     J       4o?6 

2a2  (a  +  &)g  _   2a2(a  +  6)  (a  +  6) 

a?  -  b2  4a26  (a  +  6)  (a  -  b)±a2b 

a  +  b 
2b(a  -  b)' 
>y  canceling  those  factors  which  are  common  to  both  numer- 
ator and  denominator. 

3  J  Multiply  '  a  and  ~ — -  together. 

1  J        4a3  12a  +  18     ° 

2a2  +  3a        4a2  -  6a  =  a  (2a  +  3) 2a  (2a  -  3) 
4a3  12a  +  18  4a3(2a  +  3)6 

_  2a  -  3 
12a 

4.    Multiply  2  +  ^  +  1  by -  +  --1. 
&       a  -6       a 

■+»+!=  "2  +  V  +  g»  (Art.  84), 
b       a  ab 

,  a    .    b        1        a2  -f  ft2  —  ab 

and  -  H 1  =  — -*-— . 

6       a  ao 

*+»+«»  x  g±g=?g  =  ^+*>y-^  [(3)  of  Art  41] 

a  6  a6  a~&2 

a4  +  &4  +  a2b2 


a2b2 
Otherwise  thus : 

=  ^L2  4-  i2  _l  1        a4  +  b4  +  a262 
&2       a2  a2&2 

Simplify 

^  *+i    *+j   JL-i_  ^   — i 

-      x-l       x2  -  1       (a  +  2)2  (a-l)(a!  +  2) 

^  x  +  a       \a       a;/ 


x  —  a. 


140  TO  pi  VIDE  FRACTIONS  —  EXAMPLES. 

a2W 


(•+AX6---^> 


a2  -  b 


x(a  —  x)  a  (a  +  %)  ax 

a2  -\r  2a#  +  x2       a2  —  2ax  -f  x2  a2  —  or*' 

88.  ^o  Divide  Fractions.  —  Let  it  be  required  to  divide 
a  v     c 

i  by  ? 

Denote  the  quotient  by  x.     Then,  since  the  quotient  mul- 
tiplied by  the  divisor  gives  the  dividend  (Art.  44),  we  have 

c       a 
x  X  -  =  -. 

d       b 

Multiplying  by  -,  we  have 
c 

a        c        b        c 

Therefore,  Art.  87,  and  canceling  factors  common  to  the 
numerator  and  denominator,  we  have 

ad 


X  =  . 

be 

t,,    ,  .                       a       c       ad 

That  is,                    -  -f-  -  =  —  = 
b       d        be 

ad 
bX  c 

Hence  the  following 

Rule. 
Invert  the  divisor,  and  proceed  as  in  multiplication. 


1.   Divide  a  by 

J  c 


EXAMPLES. 

b 


a  =  -  (Art,  78), 


a_^&_a       c  _  ac 
1    !    c  "~  1       6  ~  b' 


2. 


COMPLEX  FRACTIONS.  141 

ah  -  Ir    -  6 


(« 

+  by 

a-  - 

-  b2 

ab  -  &2 

b 

ah  -  b2 
{a  +  b)2 

X 

a2  - 
6 

b2 

(a  +  h)2 

a'2 

-  b2 

_  b(a.  - 

■b)(a 

+  b)(a 

-  &)  _ 

(a 

-by 

\ 

b(a  +  b)'2  a  +  b 

Simplify 
Ol    14a;2  -  7x         2x  -  1 


12a;3  +  24a;2 

a;2  -f  2a;' 

a262  +  3r/& 
4a2  -  1 

cth  -f  3 
2a  -f  1 

a2  -  121 
a2  -  4 

a  +  11 
a  +  2 

2a;2  +  13a;  +  15        2a;2 
4a;2  -  9 

-f  11a 

4a;2  - 

+  5 
1 

a;2  —  14a;  — 

15       x2  - 

■  12a;  - 

-  45 

x2  —  4a;  — 

45         a;2  - 

-  Ga;  - 

27 

Ans. 

7 
12 

ab 

2a  - 

•  1 

a  — 

11 

a  — 

2 

2x  - 

-  1 

2x  - 

3 

x  +  1 

x  +  5 


89.  Complex  Fractions.  —  A  fraction  whose  numerator 
and  denominator  are  whole  numbers  is  called  a  Simple  Frac- 
tion. A  fraction  whose  numerator  or  denominator  is  itself  a 
fraction  is  called  a  Complex  Fraction.     Thus, 

a        a 

b   a    b  ,       c      .. 

-,  _,  -  are  complex  fractions, 
c   b.  c 

c  d 

Since  a  fraction  may  he  regarded  as  representing  the 
quotient  of  the  numerator  by  the  denominator  (Art.  78), 
a  complex  fraction  may  be  regarded  in  the  same  way ; 
therefore,  to  simplify  a  complex  fraction, 

Divide  the  numerator  by  the  denominator,  as  in  division  of 
fractions  (Art.  88). 


142  EXAMPLES. 


EXAMPLES. 
1 

1.    Simplify  -,  -,  and  -. 
b   b  b 


Here  i=U*  =  lX-  =  -. 

«  b  a       a 


-  =  a-=--  =  ax6  =  ab. 
1  b 


1       a      b       a      1       a" 


The  student  should  be  able  to  write  down  the  above  results 
readily  without  the  intermediate  steps. 


Simplify %.. 


lis  fraction  =  (x  +  - \  -f-  (x  -  —\ 


This 

a;2  +  a2       re4  —  a4 


__  x2  -f  a2  a;8  .r2 

•—  — — —  x 


EXAMPLES.  143 


a2  +  b2 
Simplify 

a2  -  62 
a2  4-  &2 

wmpiiij     a  +  B 

a  —  6 

a  —  b       a  +  b 
a2  +  &2  _  g  -  b-  _  (a2  +  62)2  -  (a2  -  62)2 
a2  -  62       a2  4-  &2  (a2  -  V)  (a2  +  b2) 

_  4a262 

(a2  —  62)  (a2  -f  &2) ' 
ajj  _  a  -  &  _  (a  +  b)2  -  (a  -  b)2  _       4ab 
a  —  b       a  4-  b  a2  —  6s  a2  —  62' 

Hence  the  fraction 

4«.2fr2 4a  6 

~  (a2  -  62)  (a2  +  62)    "   a2  -  b2 

4a262  ,  a2  —  b2  ab 


(a2  —  b2)  (a2  -h  b2)  4ab  a2  +  &* 

Note  1.  —  In  this  example  the  factors  a  —  b  and  a  +  b  are  multi- 
plied together,  and  the  result  o'2  —  b2  is  used  instead  of  (a  —  b)(a  4  6). 
In  general,  however,  the  student  will  find  it  advisable  not  to  multiply 
the  factors  together  till  after  he  has  canceled  all  the  common  factors 
from  the  numerator  and  denominator. 

Note  2. — When  the  numerator  and  denominator  are  somewhat 
complicated,  to  insure  accuracy  and  neatness,  the  beginner  is  advised 
to  simplify  each  separately  as  in  the  above  example. 

Note  3.  —  The  terms  of  the  simple  fractions  which  enter  into  the 
numerator  and  denominator  of  the  complex  fraction  are  sometimes 
called  Minor  Terms.  Thus  in  Ex.  2,  a2  and  a4  are  minor  numerators, 
and  x  and  x3  are  minor  denominators. 

It  is  often  shorter  to  reduce  a  complex  fraction  to  a  simple 
one  by  multiplying  both  terms  of  the  fraction  by  the  least 
common  multiple  of  all  the  minor  denominators. 

x  +  5  +  5 

4.    Simplify  . 


1  +  ^  +  4 

x       ar 


Multiplying  both  terms  by  x2  we  get 

x3  +  5x2  4-  Gx       xjx  4-  2)(.r  4-  3)  _  x(x  j-  3) 
x2  4-  6z  4-  8    "  "    [x  4-  2)  (a  4-4)  x  4-  4 


144     A   SINGLE   FRACTION   EXPRESSED   AS   A    GROUP. 

:,.    Simplify 3x~8 


x-1 1 


4  -f  x 
In  the  case  of  Continued  Fractions,  we  begin  with  the 
lowest  complex  fraction,  and  simplify  step  by  step.     Here 
multiplying  both  terms  of  the  fraction  which  follows  x  —  1 
by  4  -}-  x,  the  given  fraction  becomes  at  once 
Sx  -  8  3a;  -  8 

.  4  +  x        =  ~        4-f-a;' 

x  —  1  —  ! a;  —  1 — — 

4  -f-  a;  —  a;  4 

and  now  multiplying  both  terms  by  4,  we  have 

4 (3a;  -  8)  _  4 (3a;  -  8)  =    4 

4  (a;  -  1)  —  (4  +  x)  "       3a;  -  8 

Simplify 

1 
a; 

6. Ans.  x  —  1. 

1  +  ± 
a; 

I  __  1.       1 

X         X2         x3  x  -f-   1 

a; 


7. 


90.  A  Single  Fraction  Expressed  as  a  Group  of 
Fractions.  —  Let  it  be  required  to  express  the  fraction 

5a;2?/  —  10a;?/2  +  15?/3  —  5a;8 
10a;2?/2 
as  a  group  of  four  fractions. 

The  fraction  =  ^L  _  1°2£  +  i&l  _   _^1_ 
10a;2?/2       10a;2?/2       10a;2?/2       lOafy2 

=  J_  _  1        3?/ a^_ 

2v/       x       2x'~       2y'z' 


EXAMPLES.  145 

Express  each  of  the   following   fractions   as  a  group  of 
simple  fractions  in  lowest  terms. 


1     3x2y  -f  xy*  —  y3 
dxy 

3       9        9a? 

a     &a*x  —  4a2ar  4-  6aa;3 
12  ax 

a2       aa;       a;2 
4     "  "s"        2  * 

o     be  +  ca  4-  ab 
«6c 

a       6       c 

EXAMPLES. 

Reduce  to  lowest  terms  the  following  examples : 
SQ  x2  +  3a?  +  2 
a;2  +  6a:  +  5* 
-£)  a?  +  10a;  +  21 
a?  _  2aj  -  15  ' 

3  3a?2  +  23a;  -  36 
4a;2  +  33a;  -  27' 

4  a;2  -  10a;  +  21 
a;3  —  46a;  —  2l'  x2  +  7x  +  3 

Here  we  see  at  once  that  the  numerator  =  {x  —  1){x  —  3);  and  we 
find  by  trial  that  z  —  1  is  a  factor  of  the  denominator. 

5.         *2  +  9*  +  20      .  Ans. 

x?  +  7x"  +  Ux  +  8 


Ajis 

X 

T 
+ 

5 

X 
X 

+ 

7 

— 

5 

3x 

— 

4 

4a; 

— 

3 

X 

— 

3 

6. 
7. 

8. 

9. 

10. 


a;3  -  10a;2  +  21a;  4-  18 
gg  -  lig  +  5 
3^  _  2x2  _  !' 

2Qq;2  +  x  -  12 
12a;3  -  5a;2  +  5a;  -  6* 
a;3  -  8x  -  3 
a;4  —  7ar  +  1 
a;3  —  x2  —  7x  +  3 


a?  4-  5 

a;2  ■+-  3a?  +  2* 

a;  +  7 

a;2  —  4a;  —  3 

6a;  —  5 

3a;2  4-  x  4-  l" 

5a;  4-  4 

3a;2  4-  x  4-  2' 

a;  —  3 

a2  -  3a?  4-  1" 

a?  —  3 

a;4  4-  2xz  4-  2a;  —  1  x2  +  1 


46 

EXAMPLES. 

11. 

x4  -  1 

X6   -    1 

12. 

gA  +  gJi  +  3.2  +  ^  +  j 

^4ws. 


10. 


x5  -  1 
6x2  -  Six  +  210 


x* 

+ 

1 

X'4 

+ 

IB3 

+  1 

1 

a; 

— 

1 

X 

- 

5 

a;3  -h  4a;2  —  47a;  —  210  t 

Reduce  to  fractional  forms  the  following  examples : 

a2  -ax  A„a  a2  -\ 

x 

,16/    a2  —  aa;  -f-  x2 


$.  a  +  x  +  a*~™.  Ans. 


1G.   x  +  5 


2a  -  15 


a:  —  3 
17.    a3  -f-  arc15  -h 


x 

a8 

— 

.VJ3 

a 

+ 

.T2 

X 

a 

a5 

3 

gt  —  ar  a'5  —  or 

Reduce  to  whole  or  mixed  quantities  the  following  examples 

18.    .  ^±?is.  x 

a  a 

1  -  2a;2 


19. 

20. 
21. 


1  -f-  x 
1  +  2a? 
1  -  3a;" 
a?  +  7 

a;  +  2* 


22.    ??  "  2 


23. 
24. 


x  +  5 
2a;2  -  7a;  -  1 
a;  —  3 
a?  -  2a;2 
a?  -  x  +  l' 


25.    **  +  1 


1  - 

-   X   — 

X2 

1    +   X 

1  -f-  bx  4-  ■ 

15a? 
L  -  3a; 

1  + 

5 
x  +  2 

3  - 

17 

x  +  5 

2a;  - 

-  1  - 

4 
a;  —  3 

SB  -    1    - 

2a; 

-  1 

a;2  - 

a;  +  1 

j-2   O-     >•   - 

i-    1     _L 

2 

a;  —  1  a?  - 

a;4  -  1 
x  +  1 


2G.    ^^  *  a?  -  x-2  +  x  -  1 


EXAMPLES.  147 

-     \ 

Perform  the  additions  and  subtractions  indicated  in  the 
following  examples : 

^y   a;  +  2/      x-  -  y2 
*Z^\      a  3a  2  ax 


a  —  x       a  -\-  x       a2  —  a2 

/Q    3 5_  _  2x  -  7 

jb       2x  —  1       4.y'2  -  1* 

30.  _2L-  +  -JL-  +    4ft2&2 


a  —  6       a  +  b       a4  —  64 
31.    _J_  _         *~3         + 


32. 


Ans.  ■ 

1 

C    -    7/' 

4a 

i 

:t  +  a; 

2x 

-  3 

x(4x2 

-  1) 

2a4  + 

Ga262 

a4  - 

-  64 

2a;2  -  9s 

+  44 

Xs  + 

64 

2a 

a?  +  4       x2  -  4a;  -f  16        xs  +  64 

a2  +  ax  +  a2  _  a;2  —  ax  -f-  a2 

a;3  —  a3  a;3  -j-  a3  a2  —  a2 

sa^i1-^-^.  i. 

SJ/         xy  xy  +  y2       x2  +  a# 

34     s2  -  2a;  +  3  g  -  2 1_  a;2-  2a; 

x3  +  1  a;2  -  a; -h  1       a;  +  1*  ar5  -f  1  ' 

35.   * » + ! .         0. 

^-^    (a;-3)(a;-4)      (»-2)(a;-4)      (a-2)(as-3) 

36.  ,       *       ;  +  — L_  +  !  1 


4(1  +  x)        4(1  -  x)        2(1  +  a;2)  1  -  a;4 

37. £z« — - — ^ -.   (See  Art.  84,  Ex.  10.) 

(a-b)(x-a)     (b-a)(b-x)     v  } 

Ans. 


38. 


(x  —  a)  (a;  —  6) 
1 


(a2-b2)(x2+b2)      (62-a2)  (a;2+a2)      (x2+a2)(x2+b2) 

Ans.  0. 


39.   1 + I -L. 

a{a  —  6)  (a  —  c)        b(b  —  a)(b  —  c)       abc 

1 


Ans. 


c(a  —  c)  (c  —  6) 


148 
40. 

41. 


+ 


EXAMPLES. 
b2 


+ 


(a-b)(a-c)       (b  —  a)(b  —  c)       (c-a)(<:-b) 


Ans.  1, 


{a-b)(a-c){x-a)      {b-a){b-c)(x-b)      (c—a)(c-b){x—c) 


Ans. 


(x  —  a)(x  —  b){x  —  c) 

Simplify  the   following  examples   in   multiplication   and 
division : 


42>    2a;2  +  5x  +  2  x        a2  +  4x 


x2  -  4 


2x2  +  9a  +  4 


Ans. 


43      2a2  -  x  -  1        4a2  +  a  -  14 

2a2  +  5a  +  2  16a2  -  49 


a  —  2 

a  -  1 

4a  -f  7* 


x. 


-^  s2+a-2       a^2  +  5a  +  4  .  /a2  +  3a+2       a  +  3\ 

J'*  a;2-a-20  ^  _x         ^-2oj-15         a2    / 

45         x*  -  8a  a2  +  2a  +  1  _^  a2  +  2a  +  4  j 

"  a2  —  4a  —  5       a3  —  a2  —  2a  a  —  5 


(a  +  ?>)2_c2  ^  ft  x    (ft-6)2-C2  1 

a2+a&  —  ac       (a-f-c)2  — 6s        ab  —  b2  —  bc  b' 

Sx       x  —  1 
2  3 


46. 


47. 


48. 


,3-(a+l)-|-  2* 


(1 


+  aa)2  _  (a  +  a)2   '   ^\1  -  a       1  +  »/ 


49.    1 


1  + 


50.    1  + 


1  +  a  + 


_2a*_ 
1  -  a 


a  +  1 

1  -f  a 
1  +  a2' 


'    \x  +  y      x-y      a2  -  y2)      \x  +  //      aJ  -  y2/ 


.     a. 


SOLUTION   OF  IIARDER   EQUATIONS  —  EXAMPLES.    149 


CHAPTER    IX. 

HARDER     SIMPLE     EQUATIONS     OF     ONE 
UNKNOWN     QUANTITY. 

91.  Solution  of  Harder  Equations.  —  We  shall  now 
give  some  simple  equations,  involving  Algebraic  fractions, 
which  are  a  little  more  difficult  than  those  in  Chapter  VI. 
These  may  be  solved,  by  help  of  the  preceding  chapter  on 
fractions,  and  by  the  same  methods  as  the  easier  equations 
given  in  Chapter  VI. 

The  following  examples  worked  in  full  will  sufficiently 
illustrate  the  most  useful  methods. 


1 .    Solve 


EXAMPLES. 

6x  -  3       Sx  -  2 


2x  +  7        x  +  5 
The  L.  C.  M.  of  the  denominators  is  (2x  -f  7)  (x  -f  5). 
Clearing  the  equation   of   fractions   by  multiplying   each 
term  by  (2x  -f  7) (as  +  5),  we  have* 

(6a;  -  S)(x  +  5)  =  (3a;  -  2)  (2a;  +  7), 
or  6x2  +  27a;  —  15  =  Qx2  +  17a;  -  14  ; 

.-.     lO.v  =  1  ;         .*.     x  —  -j^-. 
We  may  verify  this   result   by  putting   ^    for  x   in   the 
original  equation,  as  in  Chapter  VI.  ;  it  will  be  found  that 
each  member  then  becomes  — J. 

Note  1.  —  When  the  denominators  of  the  fractions  involved  con- 
tain both  simple  and  compound  factors,  it  is  frequently  best  to  multiply 
tbe  equation  by  the  simple  factors  first,  and  tben  to  collect  the  integral 
terms;  after  this  the  simplification  is  readily  completed  by  "multi- 
plying across  "  by  the  compound  factors. 

*  This  is  called  "  multiplying  across." 


150  EXAMPLES. 

2     Solve  8x  +  23  -  bx  +  2  =  2x  +  3  -  1 
20  3a;  4-  4  5 

Multiplying  by  20,  the  L.  C.  M.  of  the  simple  factors  in 
the  deuominators,  we  have 

8a;  +  23  -  2Q(5a;  +  2)  =  8a;  +  12  -  20. 
3a;  +  4 

™  •  Q1        20  (5s  +  2) 

Transposing,  31  =  — ! *-» 

3x  -f-  4 

Multiplying  across  by  3a;  4-  4,  we  have 
93a;  +  124  =  20(5a;  4-  2), 
or  84  =  7a;;         .-.     x  =  12. 

We  may  verify  this  result  as  before .;  it  will  be  found  that 
each  side  becomes  -2^. 

Note  2.  —  The  student  will  see  that,  even  when  the  denominators 
of  the  fractions  contain  all  simple  factors,  it  is  sometimes  advanta- 
geous to  clear  of  fractions  partially,  and  then  to  effect  some  reduc- 
tions, before  removing  the  remaining  fractions. 

„     c  .      x  4-  6       2a;  -  18    .   2a;  4-  3        rl    ,  3a;  4-4 
3.    solve  _ _+___=:  51  4-  — ^ 

Multiplying  by  12,  the  L.  C.  M.  of  3,  4,  12, 
12(a  +  6)  _  4 (2a;  - 18)  +  3(2»  +  3)  =  16  X  4  +  3x  +  4, 

or  12(a?  +  6)  _  gaj  +  72  +  6a;  -f  9  =  G4  4-  3a;  4-  4. 

Transposing  and  reducing,  we  have 

12(*  +  6)  -  to  -  13. 

11 

Multiplying  by  11,  we  have 

12(aj  4-  G)  ==  ll<5as  -  13), 
or  12a;  4-  72  =  55a;  -  143  ; 

.-.     43a;  =  215;         .-.     x  =  5. 
We  may  verify  this  result  as  before. 

Notk  3.  —  When  two  or  more  fractions  have  the  same  denominator, 
they  should  be  taken  together  and  simplified. 


EXAMTLES. 

4, 

13  —  °? 

,    Solve 

x  +  3 

■'+ 

23a;  +  8J  _ 
4a;  +  5 

16 

a;" 

-ft. 

+  3 

Tr 

ansposing,  we  have 

23a;  -f  8* 
4a;  -\-  5 

-4 

_  16  -  Ja 

a; 

-  13 
+  3 

4-  2a; 

en 

7a;  — 

35 

"3 

_  3  +  %x 

151 
4. 


4a;  -j-  5         X  +  3 
Multiplying  across,  we  have 

(a.  +  3)  (7a;  -  -^)  =  (3  +  lx)(Ax  +  5), 
or     7a;2  -  -3fx  +  21a;  -  35  =  12a;  +  7a;2  +  15  +  -\5-a;; 


-^-x  =  50;         .-.     a; 


600 
13T* 


c      01        a;  —  8     .a;  —  4       x  —  5    .   x  —  7 

5.    Solve = H -. 

a?  -  10       a;  —  6       x  -  7       x  —  9 

Note  4.  —  This  equation  might  be  solved  by  clearing  of  fractions, 
by  multiplying  by  the  four  denominators,  but  the  work  would  be  very 
laborious.  The  solution  will  be  much  simplified  by  transposing  two 
of  the  fractions  as  follows : 

Transposing,  we  have 

x  —  8    _  x  —  5  _  x  —  7-      x  —  4 
SB  —  10       a;  —  7  ~  x  —  9       a;  —  6* 

Simplifying  each  side  separately,  we  have 

(a?-8)  (a;-7)-(a;-5)  (s-10)  _  (a;-7)  (a;-6)-(a;-4)  (a;-9) 
(a;-10)  (a%-7)  (a- 9)  (a-6) 

or 

a2-  15a;  +  56  -  (a2-  1  ox  +  50)  _  a;2- 13a;+42-(a;2-13a;+36) 
(a;-10)  (a.--7)  (x-9)  (a?-6) 

6  6 

or  =  . 

(a;- 10)  (x-7)        (a;-9)  (aj-6) 

Dividing  by  6  and  clearing  of  fractions,  we  have 

(a-  _  9)(a;  _  6)  =  (x  -  10)  (a?  -  7), 

or  a;2  —  15a;  -J-  54  =  a;2  —  17a;  +  70 ; 

.-.     x  =  8. 


152  EXAMPLES. 

Note  5.  —  This  example  may  also  be  solved  very  neatly  by  writing 
the  equation  at  first  in  the  form 

x  —  10  +  2      x  —  6  +  2  _  x  —  7  +  2      x  —  9  -f-  2 

x  —  10  a  —  0      ~~      x  —  7  X  —  9    * 

Reducing  each  fraction  to  a  mixed  number  (Art.  81),  we  have 

1  +  — - —  +  1  +  — —  =  1  +  — £-=  +  1  +  _!L_, 
^jc-10  a  -  6  ^x-7^  x-9' 

which  gives 1 =  — — -  + 


Transposing, 


a  _  10      z  —  6      x  —  7      x  —  9 
1111 


10      x  —  7      a  —  9      x  —  6 
3  3 


(x  -  10)  (*  -  7)       (x  -  9)(a  -  6)' 
and  the  solution  may  be  completed  as  before. 

c      c<  ,       5a;  —  64       2x  —  11       4a;  —  55       a;  —  6 

6.    Solve —  = . 

a-13  jj  —  6  a-14        x  —  7 

Proceeding  as  in  the  second  method  of  Ex.  5,  we  have 

5  +  -L_  _  /2  +  _L\  =  4  +  _L_  -  (i  +  -L-) 

a; -13        \         a; -6/  a; -14       \         a;  -  7/ 

1111 


a;  —  13       a  —  6       a;  —  14       a;  —  7 
Simplifying  each  side  separately,  we  have 

7  7 


(a;  -  13)  (a-  -  6)        (a-  -  14)  (a-  -  7) 
Clearing  of  fractions,  or,  since  the  numerators  are  equal, 
the  denominators  must  be  equal,  we  have 

(x  -  13)  (a  -  6)  =  (a-  -  14)  (a-  -  7); 
.-.     a2  -  19a  +  78  =  a2  -  21a  +  98 ; 
.-.     x  =  10. 
Solve  the  following  equations : 


@ 

12          1     _  29 
a   +  12a  ~  «' 

Ans.  10. 

8. 

45                57 
2a  +  3       4a  —  5 

6. 

^ 

3a  -  1        2a  -  5       x  -  3 
2                  8                 4 

x  _ 
6  ~ 

a  +  1.           -7. 

HARDER  PROBLEMS.  153 

)    6-^t-8  -  2x  +  38  =  1.  Jns.  2. 

2a;  +  1        a;  +  12 

lTj)  A(2a;-10)— r1T(3a;-40)  =  15-J(57-x).    (See Note 2). 

Ans.  17. 

f2)^-!tJ+15-^=h!-7      (See  Note  2).        5. 
J      4         32  40  2        b      v  ' 

13.   iL±J  =   *  +  5.  6. 

3a;  -  8       3a;  -  7 

u     Gx  +  13  _   3a;  +  5   =  2a?      (gee  Note  jv  20. 

^_IV      15  5a;  -  25        5        v  ' 

<T7>  3a;  -  1        Ax  -  2        x  7 

.--V9a;  +  6        12        12a;  +  8  T 

17.— —  =— - ^L.     (See  Note  3).       1. 

a;+3      a?+l       2a;+6      2a;+2       v  ' 

18.    x_-Ll +x_-L^  =  x^LA +  x_-_2^   (See  Ex.  5).      4. 
a;  —  2      a;  —  G      a;  —  5      x  —  o 

x  —  1  _  x  —  2  _  a;  —  5  _  a;  —  6  .t 

as  —  2       a;  —  3       x  —  6       x  —  7'  *" 

6a;  +1         2a;  -  4         2a;  -  1 


15  .  7a;  -  16 


-2. 


92.  Harder  Problems  Leading  to  Simple  Equations 
■with  One  Unknown  Quantity.  —  We  shall  now  give 
some  examples  which  lead  to  simple  equations,  but  which 
differ  from  those  of  Art.  61  in  being  rather  more  difficult. 
The  statement  of  the  problem  is  rather  more  difficult  than  in 
the  examples  of  that  Article,  and  the  equations  often  involve 
more  complicated  expressions. 


154  EXAMPLES. 


EXAMPLES. 


1.  A  alone  can  do  a  piece  of  work  in  9  days,  and  B  alone 
can  do  it  in  12  days :  in  what  time  will  they  do  it  if  they 
work  together  ? 

Let    x  =  the  number  of  days  required  for  both  to  do  the 
work; 

then      -  =  the  part  that  both  can  do  in  one  day. 

x 

Also  J  =  the  part  that  A  can  do  in  one  day, 
and      y1^  =  the  part  that  B  can  do  in  one  day. 

Since  the  sum  of  the  parts  that  A  and  B  separately  can 
do  in  one  day  is  equal  to  the  part  that  both  together  can  do 
in  one  day,  we  have 

Clearing  of  fractions  by  multiplying  by  36#,  we  have 
Ax  +  3x  =  36;         .-.     x  =  5£, 

which  is  the  number  of  days  required. 

2.  A  workman  was  employed  for  60  days,  on  condition 
that  for  every  day  he  worked  he  should  receive  $3,  and  for 
every  day  he  was  absent  he  should  forfeit  $1  ;  at  the  end  of 
the  time  he  had  $48  to  receive :  required  the  number  of  days 
he  worked. 

Let  x  =  the  number  of  days  he  worked ; 

then  60  —  x  =  the  number  of  days  he  was  absent. 
Also      3x  =  the  number  of  dollars  he  received, 
and   60  —  x  =  the  number  of  dollars  he  forfeited. 
Hence,  from  the  conditions  of  the  problem,  we  have 
3a;  -  (60  -  x)  =  48. 

.-.     Ax  =  108.         .-.     x  =  27. 

That  is,  he  worked  27  days  and  was  absent  33  days. 

3.  A  starts  from  a  certain  place,  and  travels  at  the  rate 
of  7  miles  in  5  hours  ;  B  starts  from  the  same  place  8  hours 
after  A,  and  travels  in  the  same  direction  at  the  rate  of  5 


EXAMPLES.  .  155 

miles  in  3  hours  ;  how  far  will  A  travel  before  he  is  over- 
taken by  B  ? 

Let  x  =  the  number  of  hours  A  travels  before  he 

is  overtaken  ; 
then        x  —  8  =  the  number  of  hours  B  travels  before  he 
overtakes  A. 
Also  |-  =  the  part  of  a  mile  which  A  travels  in  one 

hour, 
and  f  =  the  part  of  a  mile  which  B  travels  in  one 

hour, 
Therefore  \x  =  the  number  of  miles  which  A  travels  in  x 
hours, 
and     f  (as  —  8)  =  the  number  of  miles  which  B  travels  in 
x  —  8  hours. 

Since,  when  B  overtakes  A,  they  have  traveled  the  same 
number  of  miles,  we  have  for  the  equation 

lx  =  f(*-8). 

.-.     21a  =  2bx  -  200.         .-.     x  =  50. 

Therefore  fa  =  J  X  50  =  70  miles,  the  distance  which  A 
travels  before  he  is  overtaken  by  B. 

4.  A  cistern  could  be  filled  with  water  by  means  of  one 
pipe  alone  in  6  hours,  and  by  means  of  another  pipe  alone 
in  8  hours ;  and  it  could  be  emptied  by  a  tap  in  12  hours  if 
the  two  pipes  were  closed :  in  what  time  will  the  cistern  be 
filled  if  the  pipes  and  the  tap  are  all  open? 

Let        x  =  the  required  number  of  hours. 
Then     £  =  the  part  of  the  cistern  the  first  pipe  fills  in 
one  hour ; 
therefore  -  =  the  part  of  the  cistern  the  first  pipe  fills  in 
x  hours. 
And       J  =  the  part  of  the  cistern  the  second  pipe  fills 
in  one  hour ; 
therefore  -  =  the  part  of  the  cistern  the  second  pipe  fills 
in  x  hours. 


156  .  EXAMPLES. 

Also     -jW  =  the  part  of  the  cistern  the  tap  empties  in  one 
hour ; 
therefore  —  =  the  part  of  the  cistern  the  tap  empties  in  x 
hours. 
Since  in  x  hours  the  ivhole  cistern  is  filled,  we  have,  repre- 
senting the  whole  by  unity, 

6       8  ""  12  = 
Multiplying  by  24,  we  have 

Ax  +  3a;  —  2x  =  24. 
...     a  =  4f 

5.  A  smuggler  had  a  quantity  of  brandy  which  he  ex- 
pected would  bring  him  Si 98  ;  after  he  had  sold  10  gallons 
a  revenue  officer  seized  one-third  of  the  remainder,  in  con- 
sequence of  which  the  smuggler  gets  only  $1G2:  required 
the  number  of  gallons  he  had  at  first,  and  the  price  per 
gallon. 

Let  x  =  the  number  of  gallons  ; 

198 
then  —  =  price  per  gallon  in  dollars. 

x  —  10 

— o —  =  the  number  of  gallons  seized  ; 

x  —  10       198 

and  * x  -1-  =  the  value  of  the  quantity  seized  in 

3  x 

dollars. 
Hence  we  have  the  equation 

x  ~  10  x  —  =  198  -  162  =  36. 
3  x 

Clearing  of  fractions 

66(z  -  10)  =  36a>. 

.-.     3QaJ  =  660. 

x  =  22,  the  number  of  gallons ; 

19S         198         <>a    .1  II 

anc[  =  =  $9,  the  price  per  gallon. 

x  22 


EXAMPLES.  157 

6.  A  colonel,  on  attempting  to  draw  up  his  regiment  in 
the  form  of  a  solid  square,  finds  that  he  has  31  men  over, 
and  that  he  would  require  24  men  more  in  his  regiment  in 
order  to  increase  the  side  of  the  square  by  one  man ;  how 
man}'  men  were  there  in  the  regiment? 

Let  x  =  the  number  of    men    in   the    side 

of  the  first  square  ; 
then  x2  -{-  31  =  the  number  of  men  in  the  regiment. 

Also  (x  +  l)2  —  24  =  the  number  of  men  in  the  regiment. 
Hence,  we  have  the  equation 

X2  +  3i  =  rx  +  iy  _  94, 

or  x2  +  31  =  x2  +  2x  -  23.         .-.     x  =  27. 

Hence  (27) 2  -f-  31  =  760  is  the  number  of  men  in  the 
regiment. 

Note  1.  — In  this  example  it  was  convenient  to  let  x  represent  the 
number  of  men  in  the  side  of  the  first  square  instead  of  the  number 
of  men  in  the  whole  regiment 

7.  At  the  same  time  that  the  up-train  going  at  the  rate  of 
33  miles  an  hour  passes  A,  the  down-train  going  at  the  rate 
of  21  miles  an  hour  passes  B  :  they  collide  18  miles  beyond 
the  midway  station  from  A  :  how  far  is  A  from  B  ? 

Let  x  =  the  distance  from  A  to  B  in  miles ; 

then  -  =  half  the  distance. 

2 

Also  -  +  18  =  the  number  of  miles  the  up-train  goes, 

and        -  —  18  =  the  number  of  miles  the  down-train  goefv 
2 

Now      distance  in  miles      =  fte  t[me  in  hoM8 

rate  in  miles  per  hour 

^  +  18 

Therefore  — — —  =  the  time  the  up-train  takes, 
33 

-  -  IS 
2 

and  =  the  time  the  down-train  takes. 

21 


158  EXAMPLES. 


Hence,  since  these  times  are  equal,  we  have  the  equation 

*  +  18       -  -  18 

2  2 


33  21 

Solving,  we  get  x  =  162,  which  is  the  distance  from  A  to 
B  in  miles. 

8.  A  cask,  A,  contains  12  gallons  of  wine  and  18  gallons 
of  water ;  and  another  cask,  B,  contains  9  gallons  of  wine 
and  3  gallons  of  water :  how  many  gallons  must  be  drawn 
from  each  cask  so  as  to  produce  by  their  mixture  7  gallons 
of  wine  and  7  gallons  of  water  ? 

Let  x  =  the  number  of  gallons  to  be  drawn  from  A  ; 

then  14  —  x  =  the  number  of  gallons  to  be  drawn  from  B, 
since  the  mixture  is  to  contain  14  gallons. 

Now  A  contains  30  gallons,  of  which  12  are  wine  ;  that  is, 
■J-J  of  A  is  wine.  Also  B  contains  12  gallons,  of  which  9  are 
wine  ;  that  is,  T9^  of  B  is  wine. 

Hence        \%x  =  the  number  of  gallons  of  wine  in  the 
x  gallons  drawn  from  A  ; 
and  x%(14  —  x)  =  the  number  of  gallons  of  wine  in  the 

14  —  x  gallons  drawn  from  B. 
Since  the  mixture  is  to  contain  seven  gallons  of  wine,  we 
have  1^  +  ^(14  -  X)  =  7; 

that  is,  \x  +    f(14  -  a;)  =  7. 

Solving,  we  get  x  —  10,  the  number  of  gallons  to  be  drawn 

from  A, 
and  14  —  x  =    4,  the  number  of  gallons  to  be  drawn 

from  B. 

9.  At  what  time  between  4  and  5  o'clock  is  the  minute- 
hand  of  a  watch  13  minutes  in  advance  of  the  hour-hand? 

Let  x  —  the  required  number  of  minutes  after  4  o'clock  ; 
that  is,  the  minute-hand  will  move  over  x  minute  divisions  of 
the  watch  face  in  x  minutes  ;    and  as  it  moves  12  times  as 

fast  as  the  hour-hand,  the  hour-hand  will  move  over   — 

1 2> 

minute  divisions  in  x  minutes.     At  4  o'clock  the  minute-hand 


EXAMPLES.  150 

is  20  minute  divisions  behind  the  hour-hand,  and  finally  the 
minute-hand  is  13  minute  divisions  in  advance  ;  therefore,  in 
the  x  minutes,  the  minute-hand  moves  20  -f-  13,  or  33, 
divisions  more  than  the  hour-hand. 

Hence  x  =  —  +  33  ; 

12 

therefore         lis  =  12  x  33.         .-.     x  =  3G, 

or  the  time  is  3G  minutes  past  4. 

If  the  question  be  asked,  "  At  what  times  between  4  and 

5  o'clock  will  there  be  13  minutes  between  the  two  hands?" 

we    must   also  take   into  consideration   the   case  when  the 

minute-hand  is  13  divisions  behind  the  hour-hand.     In  this 

case  the  minute-hand  gains  20  —  13,  or  7  divisions. 

Hence 


X  = 

X 

12 

+ 

7. 

11a;  = 

84. 

\     x  = 

7T'v 

times 

are 

7- 

h 

minutes 

past 

4, 

Therefore  the  times  are  7^  minutes  past  4,  and  36 
minutes  past  4. 

Note  2.  —  The  student  is  supposed  to  have  obtained  from  Arith- 
metic some  knowledge  of  ratio  and  proportion.  When  two  or  more 
unknown  quantities,  in  any  example,  have  to  each  other  a  given  ratio, 
it  is  best  to  assume  each  of  them  a  multiple  of  some  other  unknown 
quantity,  so  that  they  shall  have  to  each  other  the  given  ratio.  Thus, 
if  two  unknown  numbers  are  to  each  other  as  2  to  3,  it  is  best  to 
express  the  numbers  by  2x  and  3x,  since  these  two  numbers  are  to  each 
other  as  2  to  3.     This  will  be  illustrated  in  the  next  two  examples. 

10.  A  number  consists  of  two  digits  of  which  the  left 
digit  is  to  the  right  digit  as  2  to  3  ;  if  18  be  added  to  the 
number  the  digits  are  reversed :  what  is  the  number? 

The  student  must  remember  that  any  number  consisting 
of  two  places  of  figures  is  equal  to  ten  times  the  figure  in 
the  ten's  place  plus  the  figure  in  the  unit's  place  ;  thus,  46  is 
equal  to  10x4  +  6;  likewise  358  is  equal  to  100  x  3 
-f-  10  X  5  +  8. 

Let  2a;  =  the  left  digit ; 

then  3a;  =  the  right  digit, 

and  10  x  2a;  -f-  3a;  =  number. 


1G0  EXAMPLES. 

Hence  we  have  the  equation 

(10  x  2x  +  3a?)  +  18  =  (10  x  3a;  +  2x), 
or  20x  +  3x  +  18  =  30a  +  2x. 

x  =  2; 

.-.     2.x-  =  4,  and  3x  =  6; 
therefore  the  number  is  46. 

11.  A  hare  takes  4  leaps  to  a  greyhound's  3,  but  2  of  the 
greyhound's  leaps  are  equivalent  to  3  of  the  hare's  ;  the  hare 
has  a  start  of  50  leaps :  how  many  leaps  must  the  greyhound 
take  to  catch  the  hare? 

Let  3x  =  the  number  of  leaps  taken  by  the  greyhound  ; 
then     4x  =  the  number  of  leaps  taken  by  the  hare  in  the 

same  time. 
Also,  let  a  denote  the  number  of  feet  in  one  leap  of  the  hare  ; 
then       §a  denotes  the  number  of   feet  in  one  leap  of  the 

greyhound. 
Therefore  ox  x  §«  =  the  distance  in  3x  leaps  of  the  grey- 
hound ; 
and       (4a;  -f  50) a  =  the  distance  in  4a;  -f-  50  leaps  of  the 

hare. 
Hence  we  have  the  equation 

%ax  =  (4x  +  50)  a. 
Dividing  by  a  and  multiplying  by  2,  we  have 

9a;  =  Sx  -f  100.         .-.     x  =  100. 
Therefore  the  greyhound  must  take  300  leaps. 

Note  3.  —  It  is  often  convenient  to  introduce  an  auxiliary  symbol, 
as  a  was  introduced  in  the  above  example,  to  enable  us  to  form  the 
equation  easily;  this  can  be  removed  by  division  when  the  equation  is 
formed. 

12.  A  person  bought  a  carriage,  horse,  and  harness  for 
$600  ;  the  horse  cost  twice  as  much  as  the  harness,  and  the 
carriage  half  as  much  again  as  the  horse  and  harness :  what 
did  he  give  for  each?  Ans.  $360,  $160,  $80. 

13.  In  a  garrison  of  2711  men,  there  are  two  cavalry 
soldiers  to  twenty-five  infantry,  and  half  as  many  artillery 
as  eavalry  :   find  the  numbers  of  each.      Ans.  2150,  1%,  1)8. 


EXAMPLES.  161 

14.  A  and  B  play  for  a  stake  of  So  ;  if  A  loses  he  will 
have  as  much  as  B,  but  if  A  wins  he  will  have  three  times 
as  much  as  B  :  how  much  has  each?  Ans.  $25,  S15. 

15.  A,  B.  and  C  have  a  certain  sum  between  them;  A 
has  one-half  of  the  whole,  B  has  one-third  of  the  whole,  and 
C  has  $50  ;  how  much  have  A  and  B?  Ans.  $150,  $100. 

16.  A  number  of  troops  being  formed  into  a  solid  square, 
it  was  found  that  there  were  60  over  •  but  when  formed  into 
a  column  with  5  men  more  in  front  than  before  and  3  less  in 
depth,  there  was  just  one  man  wanting  to  complete  it :  find 
the  number  of  men.  Ans.  1504. 

17.  A  and  B  began  to  pay  their  debts  ;  A's  money  was  at 
first  f  of  B's ;  but  after  A  had  paid  $5  less  than  §  of  his 
money,  and  B  had  paid  $5  more  than  J  of  his,  it  was  found 
that  B  had  only  half  as  much  as  A  had  left :  what  sum  had 
each  at  first?  Ans.  $360,  $540. 

18.  In  a  mixture  of  copper,  lead,  and  tin,  the  copper  was 
5  lbs.  less  than  half  the  whole  quantity,  and  the  lead  and 
tin  each  5  lbs.  more  than  a  third  of  the  remainder :  find  the 
respective  quantities.  Ans.  20,  15,  15  lbs. 

19.  A  and  B  have  the  same  income  ;  A  lays  by  a  fifth  of 
his ;  but  B,  by  spending  annually  $400  more  than  A,  at  the 
end  of  four  years  finds  himself  $1100  in  debt:  what  was 
their  income?  Ans.  $625. 

20.  There  are  two  silver  cups  and  one  cover  for  both  ;  the 
first  weighs  12  ozs.,  and  with  the  cover  weighs  twice  as  much 
as  the  other  cup  without  it ;  but  the  second  with  the  cover 
weighs  a  third  as  much  again  as  the  first  without  it :  find  the 
weight  of  the  cover.  Ans.  6f  oz. 

21.  -Two  casks,  A  and  B,  contain  mixtures  of  wine  and 
water ;  in  A  the  quantity  of  wine  is  to  the  quantity  of  water 
as  4  to  3  ;  in  B  the  like  proportion  is  that  of  2  to  3.  If  A 
contain  84  gallons  what  must  B  contain,  so  that  when  the 
two  are  put  together,  the  new  mixture  may  be  half  wine  and 
half  water?  Ans.  60. 


162  EXAMPLES. 


EXAMPLES. 

Solve  the  following  equations. 

-.     x  —  2    .    ,  2x  —  1  ,,  « 

1.   -f  4  =  a; .  -a?is.  —6. 

4  3  3 

2     43;  +  17    .    Sx  ~  1Q  =  7  2. 

a;  +  3  <c  -  4 

3.   *^zA  +  («  -  l)(a?  -  2)  =  a;2  -  2a;  -  4.  7. 

o 

4     3(7  +  6g)  =  35  +  4#  j 

2  +  9a;  9  +  2x ' 

x      +  -i-  =  i.  2. 


a;  +  2       x  +  6 

„     2a?  —  5    .      a?  —  3         4a?  —  3        .,  «  K 

5  2a?  -  15  10  " 

7  4^  +  3)  =  8a;  +  37  _  7a?  -  29  6 

9  18  5a;  -  12* 

8  7  ™       =      10^ L_  -10 

a;  -  4       5a;  -  30       3a;  -  12       a?  -  6 

3a;2  _  2a;  -  8  _  (7a;  -  2)  (3a?  -  6)  2 

5  35 

11.    -?-  +  -i-  =  -*-.  iff. 

2a;  -3a;-  2       3a; +  2  2J 

19     a?  —  4  _  a;  —  5  _  x  —  7  _  a;  —  8  » 

a;  —  5        a;  —  0        x  —  8       a;  —  9 

is.  -2-  +  Ln»  =  idtl  +  °L^3.  4. 

x  —  2       x  —  1       x  —  \        x  —  G 

-  .     3  —  2a;       2a;  —  5        .              4a?2  —  1  - 

14. =  1  —  -.  —  1. 

1  —  2a;       2a;  —  7  7  —  lGa?  +  \s~ 

7  3  2  6 


EXAMPLES. 

163 

16. 

3                   30                  3                 5 
•1  -  2x       8(1  -  sb)        2  -  x       2  -  2x 

^47lS. 

-4. 

17. 

30  +  6x       CO  +  8.7;  _                 48 
x  +  1            a*  +  3                    it*  4-  1* 

3. 

18. 

a;            x  +1        a  -  8       x  —  9 

4. 

it*  —  2       a?  —  1       sc  —  6       a;  —  7 

19. 

x  4-  5       ./•  —  6       sb  —  4       a*  —  15 
x  +  4       se  —  7       a*  —  5       sb  —  16 

6. 

20. 

a-  -  7        a-  -  9        a*  -  13       x  -  15 
a;  —  9       x  —  11        x  —  15       a*  —  17 

13. 

21. 

a-  4-  3       x  +  6       SB  +  2       x  +  5 

SB  4-  6       sb  4-  9       st*  +  5       x  4-  8 

-7. 

22. 

a*  -f  2       sb  —  7       sb  +  3  _  a*  -  6 

SB                SB  —  5          SB  +    1          SB  —  4* 

2. 

23. 

4a-  -  17        IO.7;  —  13  _  8sb  -  30       5a*  — 
a*  —  4            2a*  —  3           2a*  —  7          sb  — 

4 

1 

91 

24. 

5X  _  8        6.7;  -  44        IOsb  -  8  _  sb  -  8 
sb  —  2          x  —  1           x  —  1          a;  —  6 

4. 

25. 

f«4-lYsB+2Y<B+3WrsB--lYsB-2YsB-3U3r4si 

*-2Ya* 

+1Y 

^4?is.  3. 

26.     (SB-9)(jB-7)(sB-5)(sB-l)  =  (sB-2)(iB-4)(iB-6)(SB-10). 

Ans.  5^. 


2i. 

(8SB  -  o)^a*  —  1)  =  (4a*  -  \)\\x  — 

&;. 

28. 

a-2  —  sb  4-  1        a-2  4-  sb  +  1  _  ^ 
x  —  1                 st*  4-  1 

99 

G.b  +  7        2a*  —  2        2.7*  +  1 

15            7sb  -  6             5 

30. 

,5a*  —  2  =  .25a'  4-  .2a;  —  1. 

31. 

.5a*  4-  .Ca*  —  .8  =  ,75sb  +  .25. 

32. 

,135aj  _  .225        .36        .09a*  — 

.loa*  H = 

.6                   .2                 .9 

.18 

33. 

2.b  -  3           .4a*  -  .6 

,3a*  -  .4 


.06a* 


.07 


20. 
3. 

5. 


164  EXAMPLES. 

.Sx  -  1        .5  +  1.2a? 


34. 


,5a;  -  A         2x  -  .1 


35>    (.to  -  2)  (.3a;  -  1)  _  i{Sas  _  2)  _  Ax  _  2       20< 

•  uX    —     J. 


36.    a2(x  —  a)  +  b2(x  —  b)  =  a&a.  a  +  6. 

Q7     2a;  +  3a  _  2 (3a;  +  2a) 

Dim         •  Of 

x  +  a  ox  +  a 


38.   |g  +  A  =  |g  -  lY  17a. 

s9.W*-„)-(^  =  ?(*-§}  §. 

40.  a;2  +  a(2a  -  x)  -  —  =  (x  -  5Y  +  a2#         a  +  6# 

41.  (2a-a)(a>+—  ^  =  4a/--aA-|(a-4a;)(2a  +  3a>). 


42. 
43. 
44. 


X 

— 

a 

X 

- 

b 

X2 

—  ab 

X 

— 

a 

X 

+ 

a, 

2ax 

a 

— 

b 

a 

+ 

b 

a? 

■  -  b2' 

X 

— 

a 

x  - 

-  a  - 

-  1 

2a 
Ans.  Yi 

2ab 
a  +  b 


of 

b  —  a 


x  —  b  05  —  6  —  1 


-  a  —  1       a;  —  a  —  2       x  —  6  —  1       a?  —  6  —  2 

Jns.  -J  (a  +  6  +  3). 

45.  (a?-a)8(a3+a-26)  =  (a3-6)8(a;-2a+&).       i(a+b). 

46.  3a&c+     a'ft*  (2a +  5)5^  ,^-to        _«&_ 

a  +  6       (a  +  6) 3        a  (a  +  6) 2  a         a  +  b 

47.  A  person  wishing  to  sell  a  watch  by  lottery,  charges 
$1.20  each  for  the  tickets,  by  which  he  gains  $16  ;  whereas, 
if  he  had  made  a  third  as  many  tickets  again  and  charged 
$1  each,  he  would  have  gained  one-fifth  as  many  dollars  as 
he  had  sold  tickets  :   what  was  the  value  of  the  watch? 

Ana.  $128. 


EXAMPLES.  165 

4S.  There  is  a  number  of  two  digits,  whose  difference  is 
2,  and  if  it  be  diminished  by  half  as  much  again  as  the  sum 
of  the  digits,  the  digits  will  be  reversed :  find  the  number. 

Ans.  \~>. 

49.  Find  a  number  of  3  digits,  each  greater  by  1  than 
that  which  follows  it,  so  that  its  excess  above  a  fourth  of  the 
number  formed  by  reversing  the  digits  shall  be  36  times 
the  sum  of  the  digits.  Ans.  654. 

50.  A  can  do  a  piece  of  work  in  10  days,  which  B  can  do  in 
8  ;  after  A  has  been  at  work  upon  it  3  days,  B  comes  to  help 
him  :  in  how  many  da}Ts  will  they  finish  it?        Ans.  3 J  days. 

51.  A  and  B  can  reap  a  field  together  in  7  days,  which  A 
alone  could  reap  in  10  days :  in  what  time  could  B  alone 
reap  it?  Ans.  23|  days. 

52.  A  privateer,  running  at  the  rate  of  10  miles  an  hour, 
discovers  a  ship  18  miles  off,  running  at  the  rate  of  8  miles 
an  hour :  how  many  miles  can  the  ship  run  before  it  is  over- 
taken? Ans.  72. 

53.  The  distance  between  London  and  Edinburgh  is  360 
miles ;  one  traveler  starts  from  Edinburgh  and  travels  at  the 
rate  of  30  miles  an  hour,  while  another  starts  at  the  same  time 
from  London  and  travels  at  the  rate  of  24  miles  an  hour: 
how  far  from  Edinburgh  will  they  meet?         Ans.  200  miles. 

54.  Find  two  numbers  whose  difference  is  4,  and  the  dif- 
ference of  their  squares  112.  Ans.  12,  16. 

55.  Divide  the  number  48  into  two  parts  so  that  the  excess 
of  one  part  over  20  may  be  three  tunes  the  excess  of  20  over 
the  other  part.  Ans.  32,  16. 

b(j.  A  cistern  could  be  filled  in  12  minutes  by  two  pipes 
which  run  into  it,  and  it  could  be  filled  in  20  minutes  by  one 
alone  :  in  what  time  would  it  be  filled  by  the  other  alone  ? 

Aus.  30  minutes. 
57.    Divide  the  number  90  into  four  parts  so  that  the  first 
increased  by  2,  the  second  diminished  by  2,  the  third  multi- 
plied by  2,  and  the  fourth  divided  by  2,  may  all  be  equal. 

Ans.  18,  22,  10,  40. 


166  EXAMPLES. 

58.  Divide  the  number  88  into  four  parts  so  that  the  first 
increased  by  2,  the  second  diminished  by  3,  the  third  multi- 
plied by  4,  and  the  fourth  divided  by  5,  may  all  be  equal. 

Ans.  10,  15,  3,  60. 

59.  If  20  men,  40  women,  and  50  children  receive  $500 
among  them  for  a  week's  work,  and  2  men  receive  as  much 
as  3  women  or  5  children,  what  does  each  woman  receive  for 
a  week's  work?  Ans.  85. 

60.  A  cistern  can  be  filled  in  15  minutes  by  two  pipes,  A 
and  B,  running  together ;  after  A  has  been  running  by  itself 
for  5  minutes  B  is  also  turned  on,  and  the  cistern  is  filled  in 
13  minutes  more :  in  what  time  would  it  be  filled  by  each 
pipe  separately?  Ans.  37J,  and  25  minutes. 

61.  A  man  and  his  wife  could  drink  a  cask  of  beer  in  20 
days,  the  man  drinking  half  as  much  again  as  his  wife  ;  but 
^|  of  a  gallon  having  leaked  away,  they  found  that  it  only 
lasted  them  together  for  18  days,  and  the  wife  herself  for 
two  days  longer :  how  much  did  the  cask  contain  when  full  ? 

Ans.  12  gallons. 
Let  x  =  the  number  of  gallons  the  woman  could  drink  in  a  day. 

62.  A  man,  woman,  and  child  could  reap  a  field  in  30 
hours,  the  man  doing  half  as  much  again  as  the  woman,  and 
the  woman  two-thirds  as  much  again  as  the  child  :  how  many 
hours  would  they  each  take  to  do  it  separately  ? 

Ans,  62,  93,  155. 

Let  2x  —  the  man's  number  of  hours,  Sx  =  the  woman's,  and  5x  = 
the  child's. 

63.  A  and  B  can  reap  a  field  together  in  12  hours,  A  and 
C  in  16  hours,  and  A  by  himself  in  20  hours  :  in  what  time 
(1)  could  B  and  C  together  reap  it,  and  (2)  could  A,  B,  and 
C  together  reap  it?  Ans.  21  fT  hours,  10J§  horn's. 

64.  A  can  do  half  as  much  work  as  B,  B  can  do  half  as 
much  work  as  C,  and  together  they  can  complete  a  piece  of 
wen-kin  24  days:  in  what  Lime  could  each  alone  complete 
the  work?  Ans.  168,  81,  and  42  days. 


EXAMPLES.  167 

65.  There  are  two  places  154  miles  apart,  from  which  two 
persons  start  at  the  same  time  with  a  design  to  meet ;  one 
travels  at  the  rate  of  3  miles  in  two  hours,  and  the  other  at 
the  rate  of  5  miles  in  four  hours  :  wheu  will  they  meet? 

Ans.  At  the  end  of  i~)6  hours. 

66.  Three  persons,  A,  B,  and  C,  can  together  complete  a 
piece  of  work  in  60  days  ;  and  it  is  found  that  A  does  three- 
fourths  of  what  B  does,  and  B  four-fifths  of  what  C  does : 
in  what  time  could  each  one  alone  complete  the  work? 

Ans.  240,  180,  144  days. 
Let  x  =  C's  time  of  completing  the  work,  in  days. 

67.  A  general,  on  attempting  to  draw  up  his  army  in  the 
form  of  a  solid  square,  finds  that  he  has  60  men  over,  and 
that  he  would  require  41  men  more  in  his  army  in  order  to 
increase  the  side  of  the  square  by  one  man :  how  many  men 
were  there  in  the  army  ?  Ans.  2560o 

68.  A  person  bought  a  certain  number  of  eggs,  half  of 
them  at  2  for  a  cent,  and  half  of  them  at  3  for  a  cent ;  he 
sold  them  again  at  the  rate  of  5  for  two  ceuts,  and  lost  a 
cent  by  the  bargain  :  what  was  the  number  of  eggs  ?    Ans.  60. 

69.  A  and  B  are  at  present  of  the  same  age  ;  if  A's  age 
be  increased  by  36  years,  and  B's  by  52  }'ears,  their  ages 
will  be  as  3  to  4  ;  what  is  the  present  age  of  each?   Ans.  12. 

70.  A  cistern  has  two  supply  pipes  which  will  singly  fill  it 
in  4J  hours  and  6  hours  respectively ;  and  it  has  also  a  leak 
by  which  it  would  be  emptied  in  5  hours  :  in  how  many  hours 
will  it  be  filled  when  all  are  working  together  ?         Ans.  5^. 

71.  A  person  hired  a  laborer  to  do  a  certain  work  on  the 
agreement  that  for  every  day  he  worked  he  should  receive  $2, 
but  that  for  every  day  he  was  absent  he  should  lose  $0.75  ;  he 
worked  twice  as  many  days  as  he  was  absent,  and  on  the  whole 
received  $39  :  how  many  days  did  he  work  ?  Ans.  24. 

72.  A  sum  of  mone}'  was  divided  between  A  and  B,  so 
that  the  share  of  A  was  to  that  of  B  as  5  to  3  ;  also  the 
share  of  A  exceeded  five-ninths  of  the  whole  sum  by  $200: 
what  was  the  share  of  each  person?  Ans.  $1800,  1080. 


168  EXAMPLES. 

73.  A  gentleman  left  his  whole  estate  among  his  four  sons. 
The  share  of  the  eldest  was  $-1000  less  than  half  of  the 
estate  ;  the  share  of  the  second  was  $G00  more  than  one- 
fourth  of  the  estate ;  the  third  had  half  as  much  as  the 
eldest ;  and  the  youngest  had  two-thirds  of  what  the  second 
had  :  how  much  did  each  son  receive  ? 

Ans.  $11000,  $8100,  $5500,  $5400. 
Let  x  =  the  number  of  dollars  in  the  estate. 

74.  A  and  B  shoot  by  turns  at  a  target ;  A  puts  7  bullets 
out  of  12  into  the  bull's  eye,  and  B  puts  in  9  out  of  12; 
between  them  they  put  in  32  bullets :  how  many  shots  did 
each  fire?  Ans.  24. 

75.  Two  casks,  A  and  B,  are  filled  with  two  kinds  of  sherry, 
mixed  in  the  cask  A  in  the  proportion  of  2  to  7,  and  in  the  cask 
B  in  the  proportion  of  2  to  5  :  what  quantity  must  be  taken 
from  each  to  form  a  mixture  which  shall  consist  of  2  gallons 
of  the  first  kind  and  6  of  the  second  kind?  Ans.  4|,  3 J. 

7G.  How  many  minutes  does  it  want  of  4  o'clock,  if  three- 
quarters  of  an  hour  ago  it  was  twice  as  many  minutes  past 
2  o'clock?  Ans.  25. 

Let  x  =  the  number  of  minutes  it  wants  of  4  o'clock. 

77.  At  what  time  between  3  o'clock  aud  4  o'clock  is  one 
hand  of  a  watch  exactly  in  the  direction  of  the  other  hand 
produced?  Ans.  49^  minutes  past  three. 

78.  The  hands  of  a  watch  are  at  right  angles  to  each  other 
at  3  o'clock  :  when  are  they  next  at  right  angles? 

Ans.  32T8T  minutes  past  three. 

79.  At  what  time  between  3  and  4  o'clock  is  the  minute- 
hand  one  minute  ahead  of  the  hour-hand? 

Ans.  17y\  minutes  past  three. 

80.  An  officer  can  form  his  men  into  a  hollow  square  4 
deep,  and  also  into  a  hollow  square  8  deep ;  the  front  in  the 
latter  formation  contains  16  men  fewer  than  in  the  former 
formation  :  find  the  number  of  men.  Ana.  640. 


SIMULTANEOUS   EQUATIONS.  109 


CHAPTER    X. 

SIMULTANEOUS    SIMPLE    EQUATIONS    OF   TWO 
OR    MORE    UNKNOWN    QUANTITIES. 

93.    Simultaneous   Equations   of  Two   Unknown 
Quantities.  —  If  we  have  a  single  equation  containing  two 
unknown  quantities  x  and  ?/,  we  cannot  determine  any  thing 
definite  regarding  the  values  of  x  and  y,  because  whatever 
value  we  choose  to  give  to  either  of  them,  there  will  be  a 
corresponding  value  of  the  other. 
Thus,  from  the  equation, 

2x  +  3y  =  24,   .....     .  (1) 

we  may  deduce  the  equation, 

24  -  2x 
y  =  — 3— ; 
but  we  cannot  find  the  value  of  y  from  this  equation  unless 
we  know  the  value  of  x.  We  may  give  to  x  any  value  we 
choose,  and  there  will  be  one  corresponding  value  of  y ;  and 
thus  we  may  find  as  many  pairs  of  values  as  we  please  which 
will  satisfy  the  given  equation. 

For  example,  if  x  =    3,  then  y  =  (24  —    6)  -*-  3  =  6. 
If  x  =     6,  then  y  =  (24  -  12)  -s-  3  =  4. 

3  =  2. 


3  =  -5i, 


If  a;  =     9,  then?/  =  (24  -  18) 

If  x  =  20,  then  y  =  (24  -  40) 

and  so  on. 

Any  one  of  these  pairs  of  values  f     ~~     ),  (  ),  I     ~    Y 

V/  =  <7  ^  =  4/  V/  =  2/ 

etc.,  substituted  for  x  and  ?/  in  (1)  will  satisfy  the  equation. 

Hence  a  single  equation  containing  two  unknown  quantities 

is  not  sufficient  to  determine  the  definite  value  of  either. 


170  SIMULTANEOUS  EQUATIONS. 

Suppose  we  have  a  second  equation  of  the  same  kind, 
expressing  a  different  relation  between  the  unknown  quanti- 
ties, as  for  example, 

3s  +  2y  =  26  ; (2) 

then  we  can  find  as  many  pairs  of  values  as  we  please  which 
will  satisfy  this  equation  also. 

Now  suppose  we  wish  to  determine  values  of  x  and  y 
which  will  satisfy  both  equations  (1)  and  (2)  ;  we  shall  find 
that  there  is  only  one  pair  of  values  of  x  and  y,  i.e.,  only 
one  value  of  x  and  one  value  of  y  that  will  satisfy  both 
equations.  For,  multiply  equation  (1)  by  2,  and  equation 
(2)  by  3,  and  the  equations  become 

4a.  +  Gy  =  48, (3) 

and  9x  -f-  6y  =  78 (4) 

The  coefficients  of  y  are  now  the  same  in  (3)  and  (4)  ; 
hence  if  we  subtract  each  member  of  (3)  from  the  corre- 
sponding member  of  (4) ,  we  shall  obtain  an  equation  which 
does  not  contain  y :  the  equation  will  be 

5a;  =  30 ; 
therefore  x  =  6. 

Substituting  this  value  of   x  in  either  of  the  two  given 
equations,  for  example  in  equation  (1),  we  have 
12  -f  By  =  24. 
.-.     Sy  =  12. 

•••     y  =  4. 

Thus,  if  both  equations  are  to  be  satisfied  by  the  same 
values  of  x  and  ?/,  x  must  equal  G,  and  y  must  equal  4  ;  and 

the  pair  of  values  (x  )  is  the  only  pair  of  values  which 

\y  =  4/ 

will  satisfy  both  the  given  equations. 

Simultaneous  Equations  are  those  which  are  satisfied  by 
the  sa,me  values  of  the  unknown  quantities.     Thus, 

Since  (1)  and  (2)  are  satisfied  by  the  same  values  of  a 
and  ?/,  they  are  simultaneoiis  equations. 


ELIMINATION   BY  ADDITION    OR   SUBTRACTION       171 

Independent  Equations  are  those  which  express  different 
relations  between  the  unknown  quantities,  uud  neither  can 
be  reduced  to  the  other.     Thus, 

Equations  (1)  and  (2)  are  independent,  because  they  ex- 
press different  relations  between  x  and  y.  But  2x  —  oy  =  4, 
and  8a;  —  12?/  =  1G,  are  not  independent  equations,  since 
ilie  second  is  derived  directly  from  the  first,  by  multiplying 
both  members  by  4. 

Hence  we  see  that  tiuo  independent  simultaneous  equations 
are  necessary  to  determine  the  values  of  two  unknown  quan- 
tities. 

94.  Elimination.  —  In  order  to  solve  any  two  simulta- 
neous equations  containing  two  unknown  quantities,  it  is 
necessary  to  combine  them  in  such  a  way  as  to  deduce  a 
third  equation  which  contains  only  one  of  the  unknown 
quantities ;  and  this  equation  containing  only  one  unknown 
quantity  can  be  solved  by  the  method  given  in  Chapter  IX. 
When  the  value  of  one  of  the  unknown  quantities  has  thus 
been  determined,  we  can  substitute  this  value  in  either  of 
the  given  equations,  and  then  determine  the  value  of  the 
other  unknown  quantity. 

The  process  of  combining  equations  so  as  to  get  rid  of 
either  of  the  unknown  quantities  is  coiled  Elimination.  The 
unknown  quantity  which  disappears  is  said  to  be  eliminated. 

There  are  three  methods  of  elimination  in  common  use  :  * 
(1)  by  Addition  or  Subtraction;  (2)  by  Substitution;  and 
(3)  by  Comparison. 

95.  Elimination  by   Addition   or   Subtraction. — 

1 .    Let  it  be  required  to  determine  the  values  of  x  and  y  in 
the  two  equations 

8.r  +  "ty  =  100 (1) 

12.1*  -  oy  =     88 (2) 


*  There  ie  also  a  method  by  Undetermined  Multipliers,  which  sometimes  has  the 
advantage  over  either  of  these  three  methods,  ^specially  in  Higher  Mathematics. — 
See  College  Algebra,  Art.  W,  Note. 


172      ELIMINATION  BY  ADDITION   OR  SUBTRACTION. 

If  we  wish  to  eliminate  y  we  multiply  (1)  by  5,  and  (2) 
by  7,  so  as  to  make  the  coefficients  of  y  in  both  equations 
equal.     This  gives 

40a;  +  35?/  =  500 (3) 

84x  -  35y  =  616 (4) 

Adding  (3)  and  (4),  we  have 

124a  =  1116.  .'.     x  =  9. 

To  find  y,  substitute  this  value  of  x  in  either  of  the  given 
equations.     Thus  in  (1) 

72  +  ly  =  100. 

.-.     ly=     28. 

•••     y=    4j. 

and  x  —       9  j 

In  this  solution  we  eliminated  y  by  addition. 

Otherwise  thus:  Suppose  that  in  solving  these  equations 
we  wish  to  eliminate  x  instead  of  y.  Multiply  (1 )  by  3,  and 
(2)  by  2,  so  as  to  make  the  coefficients  of  x  in  both  equations 
equal.     This  gives 

24a;  +  21?/  =  300 (5) 

24a:  -  lOy  =  176 (6) 

Subtracting  (6)  from  (5),  we  have 

31?/  =  124.         .-.     y  =  4. 

In  this  solution  we  eliminated  x  by  subtraction. 

Note  1.— The  student  will  observe  that  we  might  have  made  the 
coefficients  of  x  equal  by  multiplying  (1)  and  (2)  by  12  and  8  respec- 
tively, instead  of  by  3  and  2;  but  it  was  more  convenient  to  use  the 
smaller  multipliers,  because  it  enabled   us   to  work  with    smaller 

numbers. 

2.    Solve  2x  -h    Sy  =  31 (1) 

12x  -  17?/  =  -59 (2) 

Here  it  will  be  more  convenient  to  eliminate  a\ 


ELIMINATION   BY   ADDITION   OR   SmTRACTION.      173 

Multiplying  (1)  by  C,  to  make  the  coefficients  of  x  in  both 
equations  equal,  we  have 

12a;  4-  1%  =186;      ....   (3) 
and  from  (2)  !2z  -  I7y  =  -59 (4) 

Subtracting  (4)  from  (3),  35^  =     245. 
.-.     y  =         7. 
Substituting  this  value  of  y  in  (1),  we  have 
2x  +  21  =  31. 
.  • .     x  =     51 
and  y  =     7j 

Note  2.  — When  one  of  the  unknown  quantities  has  heen  found,  it 
is  immaterial  which  of  the  equations  we  use  to  complete  the  solution, 
though  it  is  sometimes  more  convenient  to  use  a  particular  equation 
on  account  of  its  being  less  involved  than  the  other.  Thus,  in  this 
example,  we  substituted  the  value  of  x  in  (1)  rather  than  in  (2), 
because  it  rendered  the  process  simpler. 

In  these  two  examples  we  have  eliminated  by  addition  and 
subtraction.  Hence  to  eliminate  an  unknown  quantity  by 
addition  or  subtraction,  we  have  the  following 

Rule. 

Multiply  the  given  equations,  if  necessary,  by  such  numbers 
as  ivill  make  the  coefficients  of  this  unknown  quantity  numeri- 
cally equal  in  the  resulting  equations.  Then,  if  these  equal 
coefficients  have  unlike  signs,  add  the  equations  together;  if 
they  have  the  same  sign,  subtract  one  equation  from  the  other. 

Rem. —  It  is  generally  best  to  eliminate  that  unknown  quantity 
which  has  the  smaller  coefficients  in  the  two  equations,  or  which 
requires  the  smallest  multipliers  to  make  its  coefficients  equal.  When 
the  coefficients  of  the  quantity  to  be  eliminated  are  prime  to  each 
other,  we  may  take  each  one  as  the  multiplier  of  the  other  equation. 
When  these  coefficients  are  not  prime  to  each  other,  find  their  least 
common  multiple;  and  the  smallest  multiplier  for  each  equation  will 
be  the  quotient  obtained  by  dividing  this  L.  C.  M.  by  the  coefficient  in 
that  equation.  Thus,  in  Ex.  1,  first  solution,  7  and  5  (the  coefficients 
of  y)  are  prime  to  each  other.  We  multiplied  (1)  by  5  and  (2)  by  7. 
In  the  second  solution  of  Ex.  1,  the  L.  C.  M.  of  8  and  12  (the  coeffi- 
cients of  x)  is  24;  and  henco  the  smallest  multipliers  of  (1)  and  (2) 
are  3  and  2  respectively,  which  we  used  in  that  solution. 


174     ELIMINATION  BY  ADDITION   OR  SUBTRACTION. 

3.  Solve  nix  -  213?/  =642 (1) 

114a  -  326?/  =244 (2) 

Here  we  see  that  171  and  114  contain  a  common  factor  57  ; 
so  we  shall  make  the  coefficients  of  x  in  (1)  and  (2)  equal 
to  the  least  common  multiple  of  171  and  114  if  we  multiply 
(1)  by  2  and  (2)  by  3. 

Thus,  342a  -  426?/  =  1284 

342a;  -  978?/  =     732 

Subtracting,  552,?/  =     552. 

-•■   y=  i,l 

therefore  x  =  5.  J 

Note  3.  — The  solution  is  sometimes  easily  effected  by  first  adding 
the  given  equations,  or  by  subtracting  one  from  the  other.     Thus, 

4.  Solve  127a  +    59?/  =  1928.     .     .     .   (1) 

59a  +  127?/  =  1792.     ...   (2) 

By  addition  186a  +  186?/  =  3720. 

...  x  +  y  =  20  .  .  .  .  (3) 
Subtracting  (2)  from  (1),  68a  -  G8y  =  136. 

•••     x-  y  =  2      ....   (4) 

Adding  (3)  and  (4),  2a  =  22.       .-.    a  =  11. 

Subtracting  (4)  from  (3),  2?/  =  18.       .-.    y  =     9. 

Note  4. — The  student  should  look  carefully  for  opportunities  to 
effect  such  reductions  as  are  made  in  this  example.  He  will  find  as  he 
proceeds  that  in  all  parts  of  Algebra,  particular  examples  may  be 
treated  by  methods  which  are  shorter  than  the  general  rules;  but  such 
abbreviations  can  only  be  suggested  by  experience  and  practice. 

Solve  the  following  equations  by  addition  or  subtraction  : 

5.  3a  -f-  4y  =  10,  4a  +  ?/  =  9.  Ans.  a  =  2,  ?/  =  1. 

6.  x  +  2y  =  13,  3a  +  V  =  14.  a  =  3,  y  =  5. 

7.  4a  +  ly  =  29,  a  +  By  =  11.  a  =  2,  y  =  3. 

8.  8a  —  ?/  =  34,  x  +  Hy  =  53.  a  =  5,  ?/  =  6. 

9.  14a  —  3?/  =  3!),  Gx  +  17//  =  35.  a  =  3,  ?/  =  1. 
10.  35a  -f  17/y  =  86,  56a  -  13//  =17.         x  =  1,  y  =  3. 


ELIMINATION   BY  SUBSTITUTION.  175 

11.  15a>  +  77y  =  92,  55a;  -  33y  =  22. 

^l?is.  a;  =     1,  ?/  =     1. 

12.  ox  +  2y  =  32,  20a;  —  3#  =  1.  x  =    2,  y  =  13. 

13.  7a;  +  5#  =  60,  13a;  —  11#  =  10.       a;  =    5,  y  =    5. 

14.  10a;  +  9y  =  290,  12.x-  -  11?/  =  130.    a;  =  20,  y  =  10. 

96.  Elimination  by  Substitution.  —  Find  the  values 

of  x  and  y  in  the  equations 

4a;  +  3y  =  22    .     .     .     .     .     .    (1) 

5a?  —  ly  =    6 (2) 

Transpose  3y  in  (1),  4a;  =  22  —  3y ; 

t   -i    i      j  22  -  3w 

divide  by  4,  a;  = ; 

4 

substitute  this  value  of  x  in  (2),  and  we  obtain 

,/22  -  3y\        ^  « 

5(— J— )-^=     6; 

multiply  by  4,     5(22  -  3y)  -  2%  =  24. 

.-.     t/=     2. 

Substitute  this  value  of  y  in  e#7ier  (1)  or  (2),  thus  in  (1) 
4x  +  6  =  22. 


.-.     a;  =  4) 


and 

In  this  solution  we  eliminated  x  by  substitution. 
Otherwise  thus:  from  (1)  we  have 
3y  =  22  -  4a; ; 

v    a    u    a  22  —  4a; 

divide  by  3,  ?/  = ; 

o 
substitute  this  value  of  y  in  (2), 


--ff*)-" 


multiply  by  3,      15a;  -  7(22  -  4a;)  =  18; 
that  is,  15a;  —  154  +  28a;  =  18. 

.•o    x  =    4. 


176  ELIMINATION  BY   COMPARISON. 

Substitute  this  value  of  x  in  either  (1)  or  (2),  thus  in  (1) 
16  +  %  =  22.         .-.     y  =  2. 

Here  we  eliminated  ?/  by  substitution. 

Hence,  to  eliminate  an  unknown  quantity  by  substitution, 
we  have  the  following 

Rule. 

From  either  equation,  find  the  value  of  the  unknown  quan- 
tity to  be  eliminated,  in  terms  of  the  other ;  and  substitute  this 
value  for  that  quantity  in  the  other  equation. 

Solve  by  substitution  the  following  equations : 

2.  3a;  —  Ay  =  2,  7x  —  dy  =  7.    Ans.  x  =  10,  y  =  7.' 

3.  \lx  -  ly  =  37,  Sx  +  %  =  41.        a;  =    4,  y  =  1. 

4.  6a;  —  ly  —  42,  7a;  —  6#  =  75.          x  =  21,  y  ==  12. 

5.  3a;  -  4?/  =  18,  3a;  +  2?/  =  0.            x  =     2,  y  =  -3. 

6.  4a;  -  2  =  11,  2a;  -  3?/  =  0.  a;  =    3,  y  =      2. 

7.  2a;  —  y  —  9,  3a;  —  ly  =  19.  a;  =    4,  ?/  =  —  1. 

8.  15a;  +  ly  =  29,  9a;  +  15?/  =  39.      a;  =     1,  y  =       2. 

9.  2a;  +  y  =  10,  7a;  -f  8?/  =  53.  a;  =    3,  y  =      4. 

97.  Elimination  by  Comparison.  —  Find  the  values 
of  x  and  y  in  the  equations 

2a;  +  3y  =  23     .     .     .     .     .     .   (1) 

5a?  -  2y  =  10 (2) 

Finding  the  value  of  x  in  terms  of  y  from  both  (1)  and 
(2),  we  have, 
from(l),  x=  23  ~  3y,    ......    (3) 

and  from  (2),  a;  =  10  +  2y (4) 

5 

Placing  these  two  values  of  x  equal  to  each  other,  we  have 

10  +  2?/  _  23  -  :\y 

5  2 


ELIMINATION    BY    COMPARISON.  177 

Clearing  of  fractions,  by  multiplying  by  10,  we  have 

20  -f  4y  =  115  -  Voy. 

.-.     Vdy  =     95.         .-.     y  =  5. 

Substitute  this  value  of  y  in  either  (3)  or  (4),  thus  in  (4) 

10  +  10 

x  = =  4. 

5 

In  this  solution  we  eliminated  x  by  comparison. 
Otherwise  tints:  find  the  values  of  y  in  terms  of  x  from 
(1)  and  (2). 

23  -  2x  /K. 

2/=  o ' (°) 


Therefore 


3 

53 

— 

10 

2 

5x 

— 

10 

23  -  2a; 


Clearing  of  fractions 

46  -  4a;  =  15a;  —  30. 
.:.     19a;  =  76. 

x  =     4,  the  same  as  before. 
Substituting  this  value  of  x  in  either  (5)  or  (6) ,  we  deduce 
y  =  5,  as  before. 

In  this  solution  we  eliminated  y  by  comparison. 
Hence,  to  eliminate  an  unknown  quantity  by  comparison^ 
we  have  the  following 

Rule. 

From  each  equation  find  the  value  of  the  unknown  quan- 
tity to  be  eliminated,  in  terms  of  the  other ;  then  place  these 
values  equal  to  each  other. 

Note.  — Either  of  these  methods  of  elimination  may  be  employed, 
according  to  circumstances,  and  we  shall  always  obtain  the  same 
result,  whichever  one  we  use;  each  method  has  its  advantages  in 
particular  cases.  Generally,  the  last  two  methods  introduce  fractional 
expressions,  while  the  first  method  does  not,  if  the  equations  be  first 
cleared  of  fractions.     As  a  general  rule,  the  method  by  addition  or 


178  FRACTIONAL   SIMULTANEOUS    EQUATIONS. 

subtraction  is  the  most  simple  and  elegant.  When  either  of  the  un- 
known quantities  has  1  for  its  coefficient,  the  method  by  substitution 
is  advantageous.  When  there  are  more  than  two  unknown  quantities, 
it  is  often  convenient  to  use  several  of  the  methods  in  the  same 
example. 

Solve  by  comparison  the  following  equations : 

2.  7x  —  5y  =  24,  Ax  —  oy  =  11.     Arts.  x=\l,   y  =  19. 

3.  x-  +  Sy  =  7,  ^-i^  =  3t/  -  4.  x  =    3,    y  =  2. 
3                            5 

4.  Gx  -  by  =  1,  7a;  -  4?/  =  8|.  x=    3|,  ?/  =  4. 

5.  ?L±J  +  a  =  15,  x^y  +  y  =  6.      a?  =  10,    y  =  5. 

3  5 

6.  2*  +  by  =  13,  2aj  +  4  ""  7y  =  33.       a  =19,    y  =  2. 
19  2 

7.  2aj  +  ^=-?  =  21,4v+^=i  =  29.      a?  =  10,    ?/ =  7. 

5  6 


98.  Fractional  Simultaneous  Equations  of  the 
Form  12  +  8    =  g (1) 

x        y 

21  -  12  =  3 (2) 

«       y 

If  we  cleared  these  equations  of  fractions  they  would 
involve  the  product  xy  of  the  unknown  quantities ;  and  thus 
they  would  become  quite  complex.  But  they  may  be  solved 
by  the  methods  already  given,  as  follows : 

Multiply  (1)  by  3,  —  +  —  =  24. 

x         y 

Multiply  (2)  by  2,  —  -  —  =    G. 

x         y 

Add 

Divide  by  30,  -  =     1.         ,\     x  =  3. 


90 



30. 

X 

o 

— 

1. 

X 

LITERAL   SIMULTANEOUS  EQUATIONS.  1  79 

Substitute  this  value  of  x  in  (1), 


12 

"3 

+  5  =  8. 

y 

Transpose, 

§=4. 

•••   y 

=  2. 

Solve  the  following  equations  : 

a;       y 

16. 

-<4ns. 

z  =  3,  y  =  2. 

3.   i  -  9   = 

1,^+5  = 

a;         2/ 

7. 

g  =  2,  y  =  3. 

4.    5  +  «   = 

3,  15  +   3   _ 
x         y 

4. 

a?  =  5,  y  =  3. 

5.  §  -  1  = 

3. 

»  =  2,  y  =  7. 

r     5   ,16 

b.    -  H = 

79,16-    1   = 

44. 

»  =  ii*  =  i« 

99.   Literal  Simultaneous  Equations.  —  Let  it  be 

required  to  solve 

ax     -{-by       =  c    .     .     .  .   (1) 

a'x    +  b'y      =  c'    .     .     .  .   (2) 

Multiply  (1)  by  a',         aa'x  +  a%    =  a'c.     .     .  .   (3) 

Multiply  (2)  by  a,  aa'x  -f  a&'#    =  ac' .     .     .  .   (4) 


Subtract  (3)  from 

w, 

(atf    -  a'h)y  = 

=  ac'  - 

-  a'c. 

Divide  by  {ah'  — 

a'b), 

2/  = 

ac'  - 
ah'  - 

-  a'c 

-  a'b 

To  find  the  value 

of  X 

,  eliminate  y  from  (1) 

and  (2) 

thus  : 

Multiply  (1)  by  b'  and 

(2)  by  6,    ' 
ah'x  -f  66'?/  = 

h'c    . 

t 

•  (5) 

a'bx  +  &%  = 

be'    . 

•     • 

•   (6) 

Subtract  (6)  from 

'(5), 

{ah'  —  a'h)x  = 

h'c  - 

■  be'. 

Divide  bv  (ah'  — 

aTA. 

x  = 

h'c  - 

•  be' 

ah'  —  a'b 


180   EQUATIONS    WITH   THREE    UNKNOWN  QUANTITIES. 

Solve  the  following  literal  equations  by  either  method  of 
elimination : 

2.  x  +  y  =  a  +  b,  bx  -f  ay  =  2ab.      Ans.  x  =  a,y  =  b. 

3.  (a  +  c)x  —  by  =  be,  x  +  y  =  a  +  b.       x  =  b,  y  =  a. 

,  ,        7  x  ac  be 

4.  x  -h  y  =  c,  aa?  —  by  =  c(a  —  b).      x  =    — — ,  ?/ 


5. 


a  -f  6  a  -j-  b 

?  +  |=l,|  +  2=l.       *  =  -J£_,  j,  =  A. 

a       6  b       a  a  +  b  a  +  b 

a    a4.2/-rx_^-0  x-     ah2°      v  -     a2&c 

"• 1     7    —   C'  7 —   U"  ~ "        2     i      J.2'     "  2.72* 

a      6  6      a  a2  +  &  «  -h  & 

7.    The  sum  of  two  numbers  is  a  and  their  difference  is  b  : 

find  the  numbers.  Ans.  Greater  -  +  - ;   less  -  —  -. 

Z  L  'Z  'Z 

When  the  known  quantities  in  a  problem  are  represented 
by  letters,  the  answer  furnishes  a  general  result  or  Formula 
(Art.  41)  ;  and  a  formula  expressed  in  ordinary  language, 
furnishes  a  Rule.  Thus,  in  the  present  example,  we  have 
the  following 

Rule.  The  sum  and  difference  of  two  numbers  being  given, 
to  find  the  numbers :  The  greater  number  is  equal  to  half  the 
sum  plus  half  the  difference ;  the  less  number  is  equal  to  half 
the  sum  minus  half  the  difference. 

100.  Simultaneous  Equations  with  Three  or  More 
Unknown  Quantities.  —  In  order  to  solve  simultaneous 
equations  which  contain  two  unknown  quantities,  we  have 
seen  that  we  must  have  two  equations  (Art.  93).  Similarly 
we  find  that  in  order  to  solve  simultaneous  equations  which 
contain  three  unknown  quantities,  we  must  have  three  equa- 
tions. And  generally,  when  the  values  of  several  unknown 
quantities  are  to  be  found,  it  is  necessary  to  have  as  many 
simultaneous  equations  as  there  are  unknown  quantities. 

Simultaneous  simple  equations  involving  three  or  more 
unknown  quantities,  may  be  solved  by  either  of  the  three 
methods  of  elimination  explained  in  the  preceding  articles ; 


EQUATIONS   WITH   THREE    UNKNOWN  QUANTITIES.    181 

but  the  most  convenient  method  of  elimination  is  generally 
that  by  addition  or  subtraction.  The  unknown  quantities 
are  to  be  eliminated  one  at  a  time  by  the  following 

Rule. 

If  there  he  three  simple  equations  containing  three  unknown 
quantities,  eliminate  one  of  the  unknown  quantities  from  any 
two  of  the  equations,  by  the  methods  already  explained  (Arts. 
95,  96,  97)  ;  then  eliminate  the  same  unknown  quantity  from 
the  third  given  equation  and  either  of  the  former  two ;  two 
equations  involving  two  unknown  quantities  are  thus  obtained, 
and  the  values  of  these  unknown  quantities  may  be  found  by 
the  rules  given  in  the  'preceding  Articles.  The  remaining 
unknown  quantity  may  be  found  by  substituting  these  values 
in  any  one  of  the  given  equations. 

If  four  equations  are  given  involving  four  unknown  quan- 
tities, one  of  the  unknown  quantities  must  be  eliminated 
from  three  pairs  of  the  equations.  Three  equations  involv- 
ing three  unknown  quantities  will  thus  be  obtained,  which 
may  be  solved  according  to  the  rule.  If  five  or  more  equa- 
tions are  given,  they  may  be  solved  in  a  similar  manner. 

Note  1.  —  Either  of  the  unknown  quantities  may  be  selected,  as 
the  one  to  be  first  eliminated;  but  it  is  best  to  begin  with  the  quantity 
which  has  the  simplest  coefficients;  and  when  an  unknown  quan- 
tity is  not  contained  in  all  the  given  equations,  it  is  generally  best 
to  eliminate  that  quantity  first. 

EXAMPLES. 

1.    Solve  Gx  +  '2y  —  hz  =  13, (1) 

Zx  +  3y  -  2z  =  13, (2) 

7x  +  by  —  3z  =  26 (3) 

Choose  y  as  the  first  quantity  to  be  eliminated. 
Multiply  (1)  by  3,  and  (2)  by  2, 

18s  -\-  6y  —  \bz  =  39, 
Gx  +  Gy  -     4z  =  26. 

subtracting,  12a;  -  1U  =  13 (4) 


182 


EXAMPLES. 


Multiply  (1)  by  5,  and  (3)  by  2, 

30a;  +  10y  -  25z  =  65, 
Ux  +  10?/  -     6z  =  52. 

subtracting,  16#  —  192  =13 (5) 

We  have  now  to  find  the  values  of  x  and  z  from  (4)  and  (5) . 
Multiply  (4)  by  4  and  (5)  by  3  (Art.  95,  Rem.), 

48z  -  44z  =  52, 

48z  -  57z  =  39. 


1. 


thus 

\     x  =  2. 


subtracting,  13z  =  13. 

Substitute  this  value  of  z  in  (4) 

12#  -  11  =  13. 
Substitute  these  values  of  x  and  z  in  (1)  ;  thus 
12  +  2y  -  5  =  13. 
.*.     2/=    3, 
and  z  =     1, 

as  =    2. 

Note  2.  —  Although  the  method  of  elimination  given  by  the  rule 
is  generally  the  best,  yet  in  particular  examples  solutions  may  be 
obtained  more  easily  and  elegantly  by  other  means,  which  the  student 
must  learn  by  experience.  After  a  little  practice  he  will  find  that  the 
solution  may  often  be  considerably  shortened  by  a  suitable  combination 
of  the  given  equations.     Thus,  Ex.  1  may  be  solved  as  follows : 

Add  (1)  and  (2)  and  subtract  (3), 

2x  -  4z  =  0, 

or  x  =  2z (6) 

Substitute  this  value  of  #  in  (1)  and  (2),  and  we  get 

2y  +  7z  =  13, 

3y  +  4z  =  13. 
Subtracting, 


y  -  3z  =    0. 
.-.     y  =  3z      . 
Substitute  these  values  of  x  and  y  in  (1) 
12z  +  Gz  -  52  =  13. 
.-.     z=     1; 
therefore  from  (G)  and  (7),  x  =     2, 

2/=    3. 


(7) 


thus 


EXAMPLES. 


183 


2.    Solve 


k* 


JL 

Ay 

1 

3z 

= 

i; 

1 



i 

X 

3V 

5?/  2 


Z15 


*=3, 

2 


Clearing  of  fractional  coefficients,  we  obtain 

from  (1)  -   +  - 

x        y 

from  (2)  -  -  -  =  0, 

x        y 

,         /QN  15       3    ,    60       QO 

from  (3)  1 =  32 

x        y         z 

Choose  z  as  the  first  quantity  to  be  eliminated  (Note  1) 

Multiply  (4)  by  15  and  add  the  result  to  (6)  ;  thus 

i^  +  i2=77. 

y 


(i) 

(2) 
(3) 


(5) 

(6) 


Divide  by  7, 
Multiply  (5)  by  6, 


±2  +   H  =  11 


0. 


(7) 


from  (5) 
from  (4) 

3;    Solve 


=  11. 

3, 

1, 
2. 

0, 


y 

33 
a; 

2/  = 

z  = 

5x  —  3y  —    3  =    6, (1) 

13a;  -  ly  +  3z  =  14, (2) 

7a>  —  4#  =    8 (3) 

Multiply  (1)  by  3  and  add  the  result  to  (2), 
28x  -  16y  =  32. 

Divide  by  4,  7x  —    Ay  =     8. 

Thus  we  see  that  the  combination  of  equations  (1)  and 
(2)  leads  to  an  equation  which  is  identical  with  (3)  ;  and  so 
to  find  x  and  y,  we  have  but  a  single  equation,  7x  —  Ay  =  8, 


184  EXAMPLES. 

with  two  unknown  quantities,  which  is  not  sufficient  to 
determine  the  definite  value  of  either  (Art.  93).  The 
anomaly  here  arises  from  the  fact  that  one  of  these  three 
equations  is  deducible  from  the  others  ;  in  other  words,  that 
the  three  equations  are  not  independent  (Art.  93). 

Note  3.  — Sometimes  it  is  convenient  to  use  the  following  rule: 
Express  the  values  of  two  of  the  unknown  quantities  from  two  of 
the  equations  in  terms  of  the  third  unknown  quantity,  and  substitute 
these  values  in  the  third  equation.     From  this,  the  third  unknown 
quantity  can  be  found,  and  then  the  other  two :  thus 

4.  Solve  Sx  +  4?/  -  I62  =  0, (1) 

5x  -  8y  +  10z  =  0, (2) 

2x  +  6y  +     72  =  52 (3) 

Multiply  (1)  by  2  and  add  to  (2)  ;  thus 

IIoj  —  222  =  0.  .-.     x  =  22. 

Multiply  (1)  by  5,  and  (2)  by  3,  and  subtract ;  thus 

44t/  -  1102  =  0.  .-.     y  =  —. 

Substitute  these  values  of  x  and  y  in  (3)  ;  thus 

42  +  1^2  +  72  =  52. 

.-.     2  =  2,  1 
and  x  =  4,  \ 

y  =  5.  J 

Note  4.  —  The  rule  in  Note  3  is  especially  convenient  when  all  of 
the  unknown  quantities  occur  in  only  one  equation;  thus 

5.  Solve  a,*-f?/  +  2  =  a-f6-r-c,      .     .     .     -(1) 

x  —  y  =:  b  —  a, (2) 

x  —  z  —  c  —  a (3) 

From  (2)  we  have  y  =  x  -+-  a  —  b (4) 

From  (3)  we  have  z  =  x  +  a  —  c (5) 


PROBLEMS  LEADING  TO  SIMULTANEOUS  EQUATIONS.  185 

Substitute  these  values  of  y  and  z  in  (1), 

x  -{-  x  -\-  a  —  b  +  x  +  a  —  c  =  a  +  b  -\-  c. 

.-.     3a;  =  —a  +  26  +  2c. 


from  (4) 
from  (5) 


x  =  |  (a  +  b  +  c)  —  a,  ] 
z  -  |(a  +  b  +  c)  -  e.  j 


Solve  the  following  equations  s 
G.    7a;  +  3#  —  22  =  10, 


9. 


10. 


2a;  +  5?/  +  3^  = 

39, 

5a;  —     y  -\-  oz  = 

31. 

2x  +  3y  +  ±z  = 

16, 

Sx  +  2?/  —  oz  = 

8, 

5a;  —  Gy  -f  32  = 

6. 

a:  +  2y  +  2z  = 

11, 

2a;  +     ?/  +     2  = 

7, 

3«  +  Ay  +    2  = 

14. 

x  +  3y  +  4z  = 

14, 

x  +  Zy  +    z  = 

7, 

2x  +    y  +  2z  = 

2. 

1          2  _    3  _ 

x         y         z 

1, 

5-+  *  +  «  = 

x         y         z 

24, 

7  _     8         9  _ 

SB    '        tf            8 

14. 

fa?  ==      2, 
[a  =      5. 


SB  =      3, 

2/=      2, 

2    =        1. 

a;  =      1, 

y  =     2, 

2=3. 


fa;  =  -2, 
4, 
1. 


i 

2> 


2/  = 

2  = 

X  = 

2/  = 

2  = 


101  Problems  Leading  to  Simultaneous  Equa- 
tions.—  We  shall  now  give  some  examples  of  problems 
which  lead  to  simultaneous  equations  of  the  first  degree  with 
two  or  more  unknown  quantities.  Many  of  the  problems 
given  in  Chapter  IX.  really  contain  two  or  more  unknown 
quantities,  but  the  given  relations  are  there  of  so  simple  a 
nature  that  it  is  easy  to  express  all  of  the  unknown  quanti- 


186  EXAMPLES. 

ties  in  terms  of  one  unknown  quantity,  and  thus  to  require 
but  a  single  equation.  In  the  problems  of  the  present 
chapter  the  relations  between  the  unknown  quantities  are 
not  so  simple,  and  the  solution  will  give  rise  to  simultaneous 
equations  ;  and  in  all  cases  the  conditions  of  the  problem 
must  be  sufficient  to  furnish  as  many  independent  equations 
as  there  are  unknown  quantities  to  be  determined  (Art.  100). 

EXAMPLES. 

1.  Find  two  numbers  such  that  the  greater  exceeds  twice 
the  less  by  3,  and  that  twice  the  greater  exceeds  the  less  by 
27. 

Let  x  =  the  greater  number, 
and     y  =  the  less  number. 

Then  from  the  conditions, 

x  —  2y  =    3, 
and  2x  —    y  =  27. 

Solving  these  equations,  we  have  x  =  17,  y  =  7. 

2.  If  the  numerator  of  a  fraction  be  increased  by  2  and 
the  denominator  by  1,  it  becomes  equal  to  f ;  and  if  the 
numerator  and  denominator  are  each  diminished  by  1,  it 
becomes  equal  to  \ :  find  the  fraction. 

Let  x  =  the  numerator, 
and      y  =  the  denominator. 
Then  from  the  conditions, 


x  +  2 
y  +  1 


_    5 


,  X   —    1  t 

and  =  £. 

Solving,  we  have  x  =  8,  y  ==  15. 
Hence  the  fraction  is  ^. 

3.  A  man  and  a  boy  can  do  in  15  days  a  piece  of  work 
which  would  be  done  in  2  days  by  7  men  and  i)  boys :  how 
iou£  would  it  take  one  man  to  do  it? 


EXAMPLES. 


187 


Let    x  =  the  number  of  days  in  which  one  man  would  do 
the  whole, 
and  let  y  =  the  number  of  days  in  which  one  boy  would  do 
the  whole. 

Then     -  =  the  part  that  one  man  can  do  in  one  day, 

and        -  =  the  part  that  one  boy  can  do  in  one  day. 

Then  from  the  conditions   of  the  question,  a  man  and  a 
boy  together  do  y^th  of  the  work  in  one  day ;  hence  we  have 

'£+£«* (!) 

x       y 

Also,  since  7  men  and  9  boys  do  half  the  work  in  a  day, 
we  have  -        Q 

-  +  -  =  i (2) 

x       y 

Mult: ply iug  (1)  by  9,  and  subtracting  (2)  from  it,  we  have 

!  =  *■     •••  a:  =  20- 

Thus  one  man  would  do  the  work  in  20  days. 
4.  A  railway  train  after  traveling  an  hour  is  detained  24 
minutes,  after  which  it  proceeds  at  six-fifths  of  its  former 
rate,  and  arrives  15  minutes  late.  If  the  detention  had 
taken  place  5  miles  further  on,  the  train  would  have  arrived 
2  minutes  later  than  it  did.  Find  the  original  rate  of  the 
train,  and  the  distance  traveled. 

Let  x  =  the  original  rate  of  the  train  in  miles  per 

hour ; 
and  y  =  the  number  of  miles  in  the  whole  distance 

traveled. 
Then      y  —  x  =  the  number  of  miles  to  be  traveled  after 
the  detention. 

-  =  the  number  of   hours  in  traveling  y  —  x 

miles  at  the  original  rate, 

Q  f  y    np\ 

and  -^ '-  =  the  number  of  hours  in  traveling  y  —  x 

6# 

miles  at  the  increased  rate. 


188  EXAMPLES. 

Since  the  train  is  detained  24  minutes,  and  yet  arrives  only 
15  minutes  late,  it  follows  that  the  remainder  of  the  journey 
is  performed  in  nine  minutes  less  than  it  would  have  been  if 
the  rate  had  not  been  increased  ;  hence  we  have 

S^.S^-ji        

x  bx 

If  the  detention  had  taken  place  5  miles  further  on,  there 
would  have  been  y  —  x  —  5  miles  left  to  be  traveled  after 
the  detention  ;  hence  we  have 

(2) 


or 


y  —  x  —  5 

Hy 

—  x  - 

'5)     -T?- 

X 

6x 

— '60*        •       '       * 

t  (2)  from  (1). 

5  _ 

X 

25         2 

r       —    60> 

25,   1 
47*.  f 


and  y 

5.  There  is  a  number  consisting  of  three  digits ;  the 
middle  digit  is  zero,  and  the  sum  of  the  other  digits  is  11  ; 
if  the  digits  be  reversed,  the  number  so  formed  exceeds  the 
original  number  by  495  :  find  the  number. 

Let  x  =  the  digit  in  the  unit's  place, 
and     y  =  the  digit  in  the  hundred's  place. 

Then,  since  the  digit  in  the  ten's  place  is  0,  the  number 
will  be  represented  hy  100?/  4-  x  (Art.  92,  Ex.  10)  ;  hence 
from  the  conditions,  we  have 

x  -f  y  =     11, 

and  100a;  +  y  —  (100?/  -f  x)  =  495. 

Solving,  we  get  x  =  8,  y  =  3  ;  hence  the  number  is  308. 

G.  A  sum  of  money  was  divided  equally  among  a  certain 
number  of  persons;  had  there  been  three  more,  each  would 
have  received  $1  less,  and  had  there  been  two  less,  each 
would  have  received  $1  more  than  he  did :  find  the  number 
of  persons,  and  what  each  received. 


EXAMPLES.  189 

Let     x    =  the  number  of  persons, 
and  y  =  the  number  of  dollars  which  each  received. 

Then  xy  =  the  number  of  dollars  to  be  divided,  and  from 
the  conditions,  we  have 

(0  +  8)0,-  1)  =  xy, 

and  (x-  2)(?/  +  1)  =  xy. 

Solving,  we  get  x  —  12  and  y  =  5. 

7.  A  train  traveled  a  certain  distance  at  a  uniform  rate  ; 
had  the  speed  been  6  miles  an  hour  more,  the  journey  would 
have  occupied  4  hours  less  ;  and  had  the  speed  been  6  miles 
an  hour  less,  the  jouruey  would  have  occupied  G  hours  more  : 
find  the  distance. 

Let     x    =  the  rate  of  the  train  in  miles  per  hour, 
and  y  =  the  time  of  running  the  journey  in  hours. 

Then  xy  =  the  distance  traversed,  and  from  the  conditions, 
we  have  (x  -j-  6)  (y  —  4)  =  xy, 

and  (x  -  6)  (y  +  6)  =  xy. 

Solving,  we  get  x  =  30,  and  y  =  24.     Hence  the  distance 
is  720  miles. 

8.  A,  B,  and  C  can  together  do  a  piece  of  work  in  30 
days  ;  A  and  B  can  together  do  it  in  32  days  ;  and  B  and  C 
can  together  do  it  in  120  days  :  find  the  time  in  which  each 
alone  could  do  the  work. 

Let  x  =  the  number  of  days  in  which  A  could  do  it, 
.    y  =  the  number  of  days  in  which  B  could  do  it, 
and      z  =  the  number  of  days  in  which  C  could  do  it. 
Then  we  have  from  the  conditions, 


i+i 

x       y 

+ 

1  _    i 

z 

x       y 

1 

—    "32"' 

1 
V 

+ 

z 

Solving,  we  get  x  =  40,  y  —  100,  z  =  480. 


190  EXAMPLES. 

9.  Find  the  fraction  which  is  equal  to  f  when  its 
numerator  is  increased  by  unity,  and  is  equal  to  \  when 
its  denominator  is  increased  by  unity.  Ans.  §. 

10.  A  certain  number  of  two  digits  is  equal  to  five  times 
the  sum  of  its  digits ;  and  if  nine  be  added  to  the  number 
the  digits  are  reversed  :  find  the  number.  Ans.  45. 

11.  If  15  lbs.  of  tea  and  17  lbs.  of  coffee  together  cost 
ST. 86,  and  25  lbs.  of  tea  and  13  lbs.  of  coffee  together 
cost  $10.34,  find  the  price  per  pound  of  each. 

Ans.  The  tea  cost  32  cents,  and  the  coffee  cost  18  cents, 
alb. 

12.  If  Afs  money  were  increased  by  $36  he  would  have 
three  times  as  much  as  B  ;  and  if  B's  money  were  diminished 
by  $5  he  would  have  half  as  much  as  A :  find  the  sum 
possessed  by  each.  Ans.  A  has  $42,  B  has  $26. 

13.  Find  two  numbers  such  that  half  the  first  with  a  third 
of  the  second  may  make  32,  and  that  a  fourth  of  the  first 
with  a  fifth  of  the  second  may  make  18.  Ans.  24,  60. 

14.  A  farmer  parting  with  his  stock,  sells  to  one  9  horses 
and  7  cows  for  $1200  ;  and  to  another,  at  the  same  prices, 
6  horses  and  13  cows  for  the  same  sum :  what  was  the  price 
of  each?  Ans.  $96,  $48. 

15.  Having  $45  to  give  away  among  a  certain  number  of 
persons,  I  find  that  for  a  distribution  of  $3  to  each  man  and 
$1  to  each  woman,  I  shall  have  $1  too  little;  but  that,  by 
giving  $2.50  to  each  man  and  $1.50  to  each  woman,  I  may 
distribute  the  sum  exactly :  how  many  were  there  of  men 
and  women?  Ans.  12,  10. 

16.  Find  three  numbers,  A,  B,  C,  such  that  A  with  half 
of  B,  B  with  a  third  of  C,  and  C  with  a  fourth  of  A,  may 
each  be  1000.  Ans.  640,  720,  840. 

17.  A  person  spent  $1.82  in  buying  oranges  at  the  rate  of 
3  for  two  cents,  and  apples  at  5  cents  a  dozen  ;  if  he  had 
bought  five  times  as  many  oranges  and  a  quarter  of  the 
number  of  apples  he  would  have  spent  $5.30  :  how  many  of 
each  did  he  buy?  And.  153,  192. 


EXAMPLES.  1 9 1 

EXAMPLES. 
Solve  the  following  equations  : 

1.  5x  —  ly  =  0,  7x  +  by  =  74.  ^4?is.  x  =  7,  ?/  =  5. 

2.  5x  =  7y  —  21,  21a;  —  9?/  =  75.  x  =  7,  y  =  8. 

3.  G>/  -  5a;  =  18,  12a;  -  9y  =  0.  x  =  6,  y  =  8. 

4.  7a;  +  4y  =  1,  9x  +  4?/  =  3.  x  =  1,  y  =  —If. 
o.   x  -  lly  =  1,  111?/  -  9a  =  99.  a;  =  100,  y  =  9. 

6.  8x  — 21y  =  5,6x+14y=  — 26.       x=  —  2,y=  —  1. 

7.  39a?  —  8y  =  99,  52a;  —  loy  =  80.       x  =  5,  y  =  12. 

8.  3a;  =  ly,  l'2y  =  5a;  —  1.  x  =  —  7,  ?/  =  —3. 

9.  93x  +  Iby  =  123,  15x  +  $dy  =201.       x=  l,y  =  2. 

10.  -  +  *  =  1,  -  -  %L  =  3.  x  =  4,  y  =  -3. 

2       3  4        3  J 

11.  ?-±J  +  B  =  15'  1JLir1  +  2/  =  6.        x  =  10,  2/  =  5. 

12.  f  +  f  =  34,f  +  |  =  f+12.      *  =  12,„  =  12. 

13.  lziS£  +  fci  =  2)3£±l/+         9.         ^57. 

7  o  11 

14.  |— i(2/~2)  -i(a;-3)  =0,  x-£(y-l)  -*(*-*)  =0. 

z  .4ns.  x  =  3f ,  y  =  6f . 

-K     x-2      v+2      ft  2x— 5       11  —  2?/       n       „     -     ,     0 

15. 2-L-  =  0, 2  =  0.      a;=o,  ?/  =  2. 

3  4  5  7 

16.  -  +  2  =  3x-7y— 37,3x^-7y=37.        x  =  3,y  =  -4. 

17.  (x+i)(y+5)=(x+5)(y+l),xy+x+y=(x+2)(y+2). 

^tzs.  a;  =  —2,  y  =  —2. 

18.  xy  -  (y  -  l)(x  -  1)  =  6(y -  l),x  -  y  =  1. 

vl?is.  a;  =  2-J,  y  =  1J. 

19.  L±J*  =  L+J!  =  "+■-•  +  y.         »  =  8,  »  =  16. 

3  5  7  * 


192 

20. 

21. 
22. 

23. 

24. 

25. 

26. 

27. 
28. 

29. 

31. 
32. 
33. 


EXAMPLES. 


.3x  +  .125?/  =  x  -  6,3#  -  .5?/  =  28  -  .25?/. 

^4ns.  sc  =  10,  y  =  8. 
.08z-.21?/  =  .33,  .12#  +  .7?/  =  3.54.        a=  12,^  =  3. 


?_4  =  2518  +  8=10> 
x       y  x        y 


2,5  .3 
-  +  -  =  7,- 
x       y  x 


=  11 


2/ 


^  —  TT9>  2/ 


»  + 


O            7      °  2  ( 

_   =     2'   OX —    "3 

y  y 

2x  -  -=  3, 8x  +  —  = 

2/  2/ 

a  +  ^-3 


x  =  2|,  y  =  2f. 
-19. 
a;  =  3,  y  =  6. 
_  1    7/  _  3 


+  7  =  0>3,v-10(g-l)+g=J>  +  1  =  0j 

#  —  5  6  4 

^l?is.  a;  =  4,  ?/  =  12. 


®  +  I  =  2,  bx  -  ay  =  0. 
a        6 


x  =  a,  y  =  6. 


a(a>  +  2/)  +  &(#  —  y)  =  1,  a(x  —  y)  +  b(x  +  y)  =  1. 

1 


^4ns.  a; 


-;>  2/  =  0. 


a  +  6 

x  +  ?/  =  a  H-  6,  aa;  —  6?/  +  a2  —  62  =  0. 

Ans.  x  —  26  —  a,  ?/  =  2a  —  b. 
(a+b)x  +  (a  —  b)y  =  2ac,  (6+c)a;  +  (b—c)y=  26c. 

^4?is.  x  =  y  =  c. 


x  +  2y  -  3z  =  6, 

2ic  -f  4?/  —  72;  =  9, 

Sx  —    y  —  5z  —  8. 

2x  —     y  +  2s  4, 

5x  +    y  +  3z  =  5, 

2x  -  3?/  +  4z  =  20. 

x  +  4?/  +  3z  =  17, 

3x-  +  3y  +  2  =  16, 

2a  +  2?/  +  z  =  11. 


y  =  H, 

[z  =  3. 

U  =  -1, 

j  .'/  =  -2, 

[z  =  4. 

(  x  =  2, 

y  =  3, 

I  *  =  1. 


EXAMPLES. 


19: 


34. 

2x  +  oy  +  4z  =  20, 

3x  +  Ay  +  bz  =  26, 

32  +  5#  +  Cjz  =  31. 

35. 

aj~|«=6,  y-|=8,  2- 

36. 

a       6       c       „  a    . 

1 h-  =  o,-  + 

x       y       z           x 

X  = 


Ans.  \  y 
I  J 
I* 


1, 

2, 
3. 


a=8,  y=  10,  2=14. 


2a 


y        2  a; 


-  -  =  0. 


b 

y     * 

Ans.  x  =  a,  y  =  6,  2 


37 


38. 


39. 


a      y  x      2 


1    .    1 


2,-  + 


2,1     3 


a;    y 
oy  - 


z   z     y 

1        62 


_?     I      9 


5.v;    .42  ,    - 

T  +  I  =  y  +  h 


y= 


3a;  +  1 


14  +  B 


22 

21 


+  ?■ 


40 


■I 


41. 


42. 


43. 


44. 


7a?  —  oy 
42  -  7?/ 

3u  -  2y 

2x  +  3?/ 

2a?-3#+2z=13,42M-2z=14,  ) 

4w— 2aj=30,  5?/+3z*  =  32.         J 


1.  112  -  7«  =  1, 
1,  19a;  —  3w  =  1. 

2,5./;-  72  =  11,  | 
39,  4y  +32  =  41.  j 


fa:=   4,y: 

(    2  =  16,   ?/: 

j  M=   4,  a;: 

I  y=    5,2: 


{«=    9,2 


Ix 


— 2z4-3u=17,4y— 2z+v  =  ll,  1    f  a;=   2,  y= 


I 


5y— 3a;— 2m  =  8,  4?/  —  3w  +  2r  =  9,    M  2=   3,w  = 
3z+8m=33.  j    [  v=   1. 


3a;  -  4#  -f-  32  +  3v  —  6u 
ox  -  oy  +  22  -  4w  =  11, 
10#  -  32  +  3u  -  2v  =  2, 
52  4-  4ti  4-  2t>  —  2.>;  =  3, 
L  6u  —  3v  +  4^  —  2y  =  G, 


11,  1 


a;  = 
1  *  = 


21 
To- 


2, 
3, 
1. 

9, 
25. 

12, 

7. 

1, 

5. 

4, 
3, 

2, 
1, 

3, 

-1, 

-2  = 


194  EXAMPLES. 

45.  What  fraction  is  that,  to  the  numerator  of  which  if  7 
be  added,  its  value  is  § ;  but  if  7  be  taken  from  the  denom- 
inator its  value  is  §  ?  Ans.  -^. 

46.  A  rectangular  bowling-green  having  been  measured, 
it  was  observed  that,  if  it  were  5  feet  broader  and  4  feet 
longer,  it  would  contain  11G  feet  more;  but  if  it  were  4 
feet  broader  and  5  feet  longer,  it  would  contain  113  feet 
more  :  find  its  area.  Ans.  108  sq.  ft. 

47.  A  party  was  composed  of  a  certain  number  of  men 
and  women,  and,  when  four  of  the  women  were  gone,  it  was 
observed  that  there  were  left  just  half  as  many  men  again  as 
women ;  they  came  back,  however,  with  their  husbands,  and 
now  there  were  only  a  third  as  many  men  again  as  women : 
what  were  the  original  numbers  of  each?  Ans.  12,  12. 

48.  The  sum  of  the  two  digits  of  a  certain  number  is  6 
times  their  difference,  and  the  number  itself  exceeds  6  times 
their  sum  by  3  :  find  the  number.  Ans.  75. 

49.  Divide  the  numbers  80  and  90  each  into  two  parts,  so 
that  the  sum  of  one  out  of  each  pair  may  be  100,  and  the 
difference  of  the  others  30. 

Ans.  30,  50,  and  70,  20 ;  or  60,  20,  and  40,  50. 

50.  Four  times  B's  age  exceeds  A's  age  by  20  years,  and 
one-third  of  A's  age  is  less  than  B's  age  by  2  years :  find 
their  ages.  Ans.  A  36  years,  B  14  years. 

51.  In  8  hours  A  walks  12  miles  more  than  B  does  in  7 
hours ;  and  in  13  hours  B  walks  7  miles  more  than  A  does 
in  9  hours  :  how  many  miles  does  each  walk  per  hour? 

Ans.  A  5  miles,  B  4  miles. 

52.  The  sum  and  the  difference  of  a  number  of  two  digits 
and  of  the  number  formed  by  reversing  the  digits  are  110 
and  54  respectively:  find  the  numbers.  Ans.  28,  82. 

53.  In  a  bag  containing  black  and  white  balls,  half  the 
number  of  white  is  equal  to  a  third  of  the  number  of  black  ; 
and  twice  the  whole  number  of  balls  exceeds  three  times  the 
number  of  black  balls  by  four:  how  many  balls  did  the  bag 
contain?  Ans.  8  white,  12  black. 


EXAMPLES.  195 

54.  Twenty-eight  tons  of  goods  are  to  be  carried  in  carts 
and  wagons,  and  it  is  found  that  this  will  require  15  carts  and 
12  wagons,  or  else  24  carts  and  8  wagons:  how  much  can 
each  cart  and  each  wagon  cany?  Ans.  §  tons,  f  tons. 

55.  The  first  edition  of  a  book  had  GOO  pages,  and  was 
divided  into  two  parts  ;  in  the  second  edition  one  quarter  of 
the  second  part  was  omitted  and  30  pages  added  to  the  first ; 
the  change  made  the  two  parts  of  the  same  length :  what 
were  they  in  the  first  edition?  Ans.  240,  360. 

56.  If  A  were  to  receive  $10  from  B  he  would  then  have 
twice  as  much  as  B  would  have  left ;  but  if  B  were  to  receive 
$10  from  A,  B  would  have  three  times  as  much  as  A  would 
have  left:  how  much  has  each?  Ans.  822,  $26. 

57.  A  farmer  sold  30  bushels  of  wheat  and  50  bushels  of 
barley  for  S75  ;  he  also  sold  at  the  same  prices  50  bushels 
of  wheat  and  30  bushels  of  barley  for  $77:  what  was  the 
price  of  the  wheat  per  bushel?  Ans.  $1. 

58.  A  certain  fishing  rod  consists  of  two  parts  ;  the  length 
of  the  upper  part  is  to  the  length  of  the  lower  as  5  to  7 ; 
and  9  times  the  upper  part  together  with  13  times  the  lower 
part  exceeds  11  times  the  whole  rod  by  36  inches  :  find  the 
lengths  of  the  two  parts.  Ans.  45,  63. 

59.  A  certain  company  in  a  tavern  found,  when  they  came 
to  pay  their  bill,  that  if  there  had  been  3  more  persons  to 
pay  the  same  bill,  they  would  have  paid  $1  each  less  than 
they  did  ;  and  if  there  had  been  2  fewer  persons  they  would 
have  paid  $1  each  more  than  they  did :  find  the  number 
of  persons,  and  the  number  of  dollars  each  paid. 

Ans.  12,  5. 

60.  There  is  a  rectangular  floor,  such  that  if  it  had  been 
2  feet  broader,  and  3  feet  longer,  it  would  have  been  64 
square  feet  larger ;  but  if  it  had  been  3  feet  broader,  and  2 
feet  longer,  it  would  have  been  68  square  feet  larger :  find 
the  length  and  breadth  of  the  floor.  Ans.  14  ft.,  10  ft. 

Let  x  =  the  length,  and  y  =  the  breadth,  of  the  floor  in  feet;  then 
xy  =  the  surface  of  the  floor  in  square  feet. 


196  EXAMPLES. 

61.  When  a  certain  number  of  two  digits  is  doubled,  and 
increased  by  36,  the  result  is  the  same  as  if  the  number  had 
been  reversed,  and  doubled,  and  then  diminished  by  36 ; 
also  the  number  itself  exceeds  4  times  the  sum  of  its  digits 
by  3  :  find  the  number.  Ans.  59. 

62.  Two  passengers  have  together  560  lbs.  of  luggage, 
and  are  charged  for  the  excess  above  the  weight  allowed  62 
cents  and  $1.18  respectively  ;  if  the  luggage  had  all  belonged 
to  one  of  them  he  would  have  been  charged  $2.30 :  how 
much  luggage  is  each  passenger  allowed  without  charge  ? 

Ans.  100  lbs. 

63.  A  farmer  has  28  bushels  of  barley  at  56  cents  a 
bushel ;  with  these  he  wishes  to  mix  rye  at  72  cents  a  bushel, 
and  wheat  at  96  cents  a  bushel,  so  that  the  mixture  may 
consist  of  100  bushels,  and  be  worth  80  cents  a  bushel :  how 
many  bushels  of  rye  and  wheat  must  he  take?     Ans.  20,  52. 

64.  A  and  B  ran  a  race  which  lasted  5  minutes  ;  B  had  a 
start  of  20  3*ards  ;  but  A  rau  3  }-ards  while  B  was  running  2, 
and  won  by  30  yards  :  find  the  length  of  the  course  and  the 
rate  of  each  per  minute. 

Ans.  150  yards,  30  yards,  20  yards. 

65.  A  and  B  can  together  do  a  certain  work  in  30  days ; 
at  the  end  of  18  days  however  B  is  called  off  and  A  finishes 
it  alone  in  20  days  more :  find  the  time  in  which  each  could 
do  the  work  alone.  Ans.  50,  75. 

66.  A,  B,  and  C  can  together  drink  a  cask  of  beer  in  15 
days ;  A  and  B  together  drink  four-thirds  of  what  C  does  ; 
and  C  drinks  twice  as  much  as  A :  find  the  time  in  which 
each  alone  could  drink  the  cask  of  beer.        Ans.  70.  42,  35. 

67.  A  and  B  run  a  mile  ;  at  the  first  heat  A  gives  B  a 
start  of  20  yards,  and  beats  him  by  30  seconds ;  at  the 
second  heat  A  gives  B  a  start  of  32  seconds,  and  beats  him 
by  9T5T  yards :  find  the  rate  per  hour  at  which  A  runs. 

^ins.  12  miles. 

68.  A  and  B  are  two  towns  situated  24  miles  apart,  on 
the  same  bank  of  a  river.     A  man  goes  from  A  to  B  in  7 


EXAMPLES.  197 

hours,  by  rowing  the  first  half  of  the  distance,  and  walking 
the  second  half.  In  returning  he  walks  the  first  half  at 
three-fourths  of  his  former  rate,  but  the  stream  being  with 
him  he  rows  at  double  his  rate  in  going  ;  and  he  accomplishes 
the  whole  distance  in  G  hours.  Find  his  rates  of  walking  and 
rowing  up  stream.  Ans.  4  miles  walking,  3  miles  rowing. 
G9.  A  railway  train  after  traveling  an  hour  is  detained  15 
minutes,  after  which  it  proceeds  at  three-fourths  of  its  former 
rate,  and  arrives  24  minutes  late.  If  the  detention  had 
taken  place  5  miles  further  on,  the  train  would  have  arrived 
3  minutes  sooner  than  it  did.  Find  the  original  rate  of  the 
train  and  the  distance  traveled. 

Ans.  33^  miles  per  hour,  48J  distance. 

70.  The  time  which  an  express  train  takes  to  travel  a 
journey  of  120  miles  is  to  that  taken  by  an  ordinaiy  train  as 
9  to  14.  The  ordinary  train  loses  as  much  time  in  stopping 
as  it  would  take  to  travel  20  miles  without  stopping.  The 
express  train  loses  only  half  as  much  time  in  stopping  as  the 
ordinary  train,  and  it  also  travels  15  miles  an  hour  faster. 
Find  the  rate  of  each  train.  Ans.  45,  30  miles  per  hour. 

71.  A  and  B  can  perform  a  piece  of  work  together  in  48 
days ;  A  and  C  in  30  days  ;  and  B  and  C  in  26§  days :  find 
the  time  in  which  each  could  perform  the  work  alone. 

Ans.  120,  80,  40  days. 

72.  There  is  a  certain  number  of  three  digits  which  is 
equal  to  48  times  the  sum  of  its  digits  ;  and  if  198  be  sub- 
tracted from  the  number  the  digits  will  be  reversed  ;  also  the 
sum  of  the  extreme  digits  is  equal  to  twice  the  middle  digit : 
find  the  number.  Ans.  432. 

73.  A  man  bought  10  horses,  120  oxen,  and  46  cows. 
The  price  of  3  oxen  is  equal  to  that  of  5  cows.  A  horse, 
an  ox,  and  a  cow  together  cost  a  number  of  dollars  greater 
by  300  than  the  whole  number  of  animals  bought ;  and  the 
whole  sum  spent  was  $93GG.  Find  the  price  of  a  horse,  an 
ox,  and  a  cow  respectively.  Ans.  $420,  $35,  $21 


198  INVOLUTION   OF  POWERS    OF  MONOMIALS. 


CHAPTER     XI. 

INVOLUTION  AND  EVOLUTION. 

102.  Involution  is  the  process  of  raising  an  expression 
to  any  required  power.  Involution  is  therefore  only  a 
particular  case  of  multiplication,  in  which  the  factors  are 
equal  (Art.  12).  It  is  convenient,  however,  to  give  some 
rules  for  writing  down  the  power  at  once. 

It  is  evident  from  the  Rule  of  Signs  (Art.  36)  that, 

(1)  No  even  power  of  any  quantity  can  be  negative. 

(2)  Any  odd  power  of  a  quantity  will  have  the  same  sign 
as  the  quantity  itself.     Thus, 

(-a)2  =  (_a)  (-a)  =  +a2, 
(_a)3  =  (-a)(-u)(-a)  =  +a\-a)  =  -a3, 
(_a)4  =  (-a)(-a)(-a)(-a)  =  (-a3)(-a)  =  -f-a4; 
and  so  on. 

Note. —  The  square  of  every  expression,  whether  positive  or 
negative,  is  positive. 

103.  Involution  of  Powers  of  Monomials.  —  From 
the  definition,  we  have,  by  the  rules  of  multiplication, 

(a2)8=(a2)(a2)(a2)  =  a2  +  2+2  =  a6. 
(_a3)2=(-a3)(-«3)=a3  +  3  =  a6. 
(_3a3)2=(-3a3)(-3a3)  =  (-  3)2(a3)2  =  9a6. 
Generally, 

(a"')"  =  am  •  am  •  am  •  am  .  .  .  .  to  n  factors 

_am  +  m+m+m £Q  n  termg 

=  amn. 

(ab)m=ab  •  a&  •  ab torn  factors 

=  (aaaa  ....  to m  factors)  X  {bbbb  ....  torn  factors 
=  a"lbm. 
Hence  Cab)"1  =  a"'6,n, 


EXAMPLES. 


199 


and  so  on  for  any  number  of  factors.     Thus,  the  mth  power 
of  a  product  is  equal  to  the  product  of  the  mtk  powers  of  its 
factors. 
Hence  (axW<? )m  =  (ax)m  (bv)m  (cz)m 

_    axmfyjmczm 

Hence,  to  raise  any  power  of  a  quantity  to  any  other 
power,  we  have  the  following 

Rule. 

liaise  the  numerical  coefficient  to  the  required  power  by 
Arithmetic,  multiply  the  exponent  of  each  factor  by  the 
exponent  of  the  required  power,  and  give  the  proper  sign 
to  the  result. 


V 

71      a       a       a               ,         »    . 

=  -  X  -  X  - torn  factors 

b       b       b 

aaa  ....  to  m  factors 

bbb  .  ...  to  7n  factors 

_  am 
-  p.' 

Hence,  for  obtaining  any  power  of  a  fraction  :  liaise  both 
the  numerator  and  denominator  to  that  power,  and  give  the 
proper  sign  to  the  result. 


EXAMPLES 

Show  that 

1.  (a2b3)2  =  a*b\ 

2.  (-a2b3)3=  -a%\ 

3.  (aW)4  =  a8612c16. 

4.  (-2x2)5  =  -32z10. 

5.  (-3a63)6  =  729a6618. 

6. 


/a_*\8      a6 
\b2)       F 

\     be2)  "        bbc10' 


bi-2ci8- 


8  (-*)' 

'•■(-S)"- 

ii.  /   xyzy=xY 


27a;15 
125a;9 


200  INVOLUTION   OF  BINOMIALS. 

104.  Involution  of  Binomials.  —  We  have  already 
proved  by  actual  multiplication  (Art.  41),  the  two  following 
cases  of  the  involution  of  binomial  expressions  : 

(a  +  b)2  =  a2  +  2ab  +  b2    .     .     .     .   (1) 

(a  -  b)2  =  a2  -  2ab  +  b2    .     .     .     .   (2) 

If  we  multiply  (1)  and  (2)  by  a  +  b  and  a  —  b  respec- 
tively, we  have 

(a  +  b)3  =  a3  +  Serb  +  Sab2  +  b5  .     .     .   (3) 

(a  -  b)3  =  a3  -  Serb  +  Sab2  -  b3 .     .     .   (4) 

If  we  multiply  (3)  and  (4)  by  a  -f  b  and  a  —  b  respec- 
tively, we  shall  have 

(a  +  6)4  =  a4  +  4a3b  +  6a2b2  +  4a&8  +  &4- 

(a  -  6)4  =  a4  -  4a86  +  6a262  -  4a&8  +  64. 

Ify  multiplying  these  two  results  by  a  -f-  6  and  a  —  b 
respectively  we  should  obtain  {a  -f  b)5  and  (a  —  6)5;  and 
by  continuing  the  process  we  could  obtain  any  required 
power  of  (a  +  b)  or  (a  —  6).     Hence  the  following 

Rule. 

Multiply  the  binomial  by  itself,  until  it  has  been  taken  as  a 
factor  as  many  times  as  there  are  units  in  the  exponent  of  the 
required  power. 

This  rule,  however,  would  be  very  laborious  in  finding  any 
high  power,  for  instance  (a  +  b)'20.  In  Chapter  XVII  we 
shall  prove  a  theorem,  called  the  Binomial  Theorem,  by  the 
aid  of  which  any  power  of  a  binomial  expression  can  be 
obtained  without  the  labor  of  actual  multiplication. 

Since  the  above  formulae  are  true  for  all  values  of  a  and  7>, 
we  can  write  down  the  squares  and  the  cubes  of  any  binomial 
expressions.     Thus, 

1.    (a4  -  b<)2  =  (a*)*  +  2(a4)(-&4)  +  (-biy 

=  a8  -  2a464  +  b*. 


INVOLUTION   OF  POLYNOMIALS.  201 

Show  that 

2.  (2a;  +  Zyf  =  4a-2  +  I2xy  +  9t/2. 

3.  (3a;  +  5#)'2  =  Oar  +  30a;?/  -f  2oy2. 

4.  (aj  -  2y)3  =  a;3  -  Ga%  +  12a;?/2  -  8y\ 

5.  (2a6  -  3c)3  =  8a363  -  36a262c  +  oiabc2  -  27c3. 

6.  (5a2  -  362)3  =  125a6  -  225a462  +  135a264  -  2766. 

105.    Involution   of   Polynomials.  —  We  may  now 

apply  the  formulae  of  Art.  104  to  obtain  the  powers  of  any 
trinomial  or  polynomial.     Thus  from  (1) 
(a  +  b  +  cy=  [(a  +  b)  +  cy 

==  (a  +  b)2  +  2(a  -f  b)c  +  c2 

=  a2  +  b2  +  c2  +  2a6  +  2ac  -f-  26c     .   (1) 

In  the  same  way  we  may  prove 

(a+b+c+dy2=a2+b2+c2+d2+2ab+2ac+2ad+2bc+2bd+2cd    .    (2) 

We  observe  in  both  (1)  and  (2)  that  the  square  consists  of 

(1)  the  sum  of  the  squares  of  the  several  terms  of  the 
given  expressions ; 

(2)  twice  the  sum  of  the  products  two  and  two  of  the 
several  terms,  taken  with  their  proper  signs. 

The  same  law  holds  whatever  be  the  number  of  terms  in 
the  expression  to  be  squared.     Hence  the  following 

Rule. 

To  find  the  square  of  any  polynomial,  write  the  square  of 
each  term  together  ivith  twice  the  product  of  each  term  by  each 
of  the  terms  following  it. 

From  (3)  of  Art.  104  we  obtain  the  cube  of  a  trinomial  as 
follows : 
(a+b+cf  =  [a+(&+c)]3 

=  a8+3a2(6+c)+3a(6+c)2+(6+c)a 

=  a3+63+c3+3a2(6  +  c)+36^  +  c)+3c2(a+6)+6a&c      .  (3) 


202  INVOLUTION   OF  POLYNOMIALS. 

Hence  to  find  the  cube  of  a  trinomial  we  have  the  following 

Rule. 

Write  the  cube  of  each  term,  together  ivith  three  times  the 
product  of  the  square  of  each  term  by  the  sum  of  the  other 
two,  and  six  times  the  product  of  the  three  terms. 

Formulae  (1),  (2),  and  (3)  may  be  used  for  obtaining  the 
squares  and  cubes  of  any  polynomial  expressions,  as  ex- 
plained in  Art.  104.  Thus,  if  we  require  (1  —  2x  -f  3a;2)2, 
in  formula  (1)  we  put  1  for  a,  —2x  for  6,  and  3a;2  for  c,  and 
obtain 

1.  (l-2x+3x2)2 

=  (l)2+(-2a;)2+(3^)2+2(l)(-2.T)+2(l)(3rr2)+2(-2a;)(3a;2) 
=  1  +  \x-+  9aJ*-4a+  6x2- 1 2a;3 
=  l-4a;+10a;2-12a;3+9a;4. 

Similarly  by  (3)  we  have 

2.  (l-2z+3a;2)3 

=  (l)3+(_2o;)3+(3a;2)3+3(l)2(-2a;+3a;2)-}-3(-2a;)2(l+3arQ) 

+3(3a;2)2(l-2a;)+G(l)(-2a;)(3a;2) 

=  l_8a;3+27a;6+3(-2a;+3a;2)  +  12a;2(l+3a;2) 

+27a!*(l-2aB)-36a^ 

=  l_6a;+21a;2-44a;3+G3a;4-54a;5+27a;6. 

Show  that 

3.  (1  —  x  +  x1)2  =  1  -  2x  +  3a;2  -  2a;3  -f-  x*. 

4.  (1  4  $x+2x*)*  =  1  4  Gx  +  13a;2  4-  12a;3  +  4aA 

5.  (*  -  26  4  -Y  =  -  +  46a  4  —  -  2ab  +  —  -  be. 
\2  4/         4  1G  4 

6.  (|^  _  x  +  |)i  =  il_  _  *£  +  3aji  _  3a.  +  J. 

J  o 

7.  (1  4-  &  4-  a;2)3  =  1  4  3a  +  Gar  4  7a-3  4-  Gx4  4  3a-5  -f  aj6. 

8.  (1  +  x  —  ar)3  =  1  4-  ox  -  r>-;i 


EVOLUTION    OF  MONOMIALS.  203 

EVOLUTION. 

106.  Evolution  —  Evolution  of  Monomials.  —  Evo- 
lution is  the  operation  of  finding  any  required  root  of  a 
number  or  expression.  A  root  of  any  quantity  is  a  factor 
which  being  multiplied  by  itself  a  certain  number  of  times 
produces  the  given  quantity  (Art.  13).  Hence  Evolution  is 
the  inverse  of  Involution  (Art.  102). 

The  symbol  which  denotes  that  a  square  root  is  to  be 
extracted  is  \J~  ;  and  for  other  roots  the  same  symbol  is 
used,  but  with  a  figure  called  the  index  written  above  to 
indicate  the  root  (Art.  13). 

By  the  Rule  of  Signs  (Art.  36),  we  see  that 

(1)  any  even  root  of  a  positive  quantity  may  be  either 
positive  or  negative; 

(2)  every  odd  root  of  a  quantity  has  the  same  sign  as  the 
quantity; 

(3)  there  can  be  no  even  root  of  a  negative  quantity. 
Thus,  (l)a  x  a  =  a'2,  and  (  — «)(  —  «)=  a%  >  therefore  there 
are  two  roots  of  a2,  namely,  -fa  and  —a. 

(2)  (  —  «)(  —  «)(  —  a)  =  —  a3;  therefore  the  cube  root 
of  —  a3  is  —a. 

(3)  There  can  be  no  square  root  of  —a2 ;  for  if  any 
quantity  be  multiplied  by  itself,  the  result  is  a  positive 
quantity. 

There  can  be  no  even  root  of  a  negative  quantity,  because 
no  quantity  raised  to  an  even  power  can  produce  a  negative 
result.  Even  roots  are  called  impossible  roots  or  imaginary 
roots. 

Since  the  nih  power  of  am  is  amn  (Art.  103),  it  follows  that 
the  n*  root  of  amn  is  am. 

Also,  the  mth  power  of  a  product  is  the  product  of  the  mth 
powders  of  its  factors  (Art.  103)  ;  hence,  conversely,  the 
mth  root  of  a  product  is  the  product  of  the  mth  roots  of  its 
factors.     Thus, 

>Jabc  =  v'a  ft  sjc ;  1/ab  =  7a  V&. 


204  EVOLUTION   OF  MONOMIALS. 

Again,  we  have  (Art.  103) 

(cfV'c2  .  .  .  .  )m  =  a™  bym  c™  .... ; 


therefore,  conversely,  n\Jax,Hb*m<fm  .  .  .  .  =  ax&<f 

Hence  to  extract  any  root  of  a  monomial,  we  have  the 
following 

Rule. 

Extract  the  required  root  of  the  coefficient  by  Arithmetic, 
then  divide  the  exponent  of  every  factor  in  the  expression  by 
the  index  of  the  root,  and  give  the  proper  sign  to  the  result. 

Thus,  for  example, 

Vo1  =  a2 ;     sfa1!?  =  a3b2 ;     V^  =  -x3 ;     y/x™  =  x2 ; 


Vl6a2&4  =  ±ab2',     V-8aW2  =  -2a2b3c*. 

To  obtain  any  root  of  a  fraction :  Find  the  root  of  the 
numerator  and  denominator,  and  give  the  proper  sign  to 
the  result. 

This  is  the  converse  of  the  rule  in  Art.  103. 


^  ,     4s/  — 27a6  3a2 

i  or  example,  \    ■ = . 

1     'V       6463  46 

Note  1.  — Since  every  positive  quantity  has  two  square  roots  equal 
in  magnitude  but  opposite  in  sign,  it  is  customary  to  prefix  the  double 
sign  ±,  read  plus  or  minus,  to  a  quantity  when  we  wish  to  indicate 
that  it  is  either  +  or  — .     Thus 


y256ajy  =  {/SbY  =  ±4xf. 

Note  2. — Any  quantity  whose  root  can  be  extracted  is  called  a 
perfect  power.  When  the  square  root  of  an  expression  which  is  not 
a  perfect  square,  or  the  cube  root  of  an  expression  which  is  not  a 
perfect  cube,  is  required,  the  operation  cannot  be  performed.  Thus 
we  cannot  take  the  cube  root  of  a-  since  the  exponent  2  is  not  divisible 
by  the  index  3.  At  present  we  can  only  express  the  result  thus  (*/a2. 
Also,  ya,  \a?,  ya6,  cannot  at  present  be  otherwise  expressed  ;  and 
similarly  in  other  cases.  Such  quanl  it  ies  are  called  nurds  or  irrational 
quantities,  and  will  be  considered  in  Chapter  XII. 


SQUARE   ROOT   OF  A   POLYNOMIAL.  205 

EXAMPLES. 


Show  that 


1.    \aHrci2  =  ±a46c6 


2.    \04xYs  =   ±8 


2-a 


'   V    81a;10  9z5    ' 

4.    V^ToW  =  Scrbc3 ; 


5-    V-343a12618  =  -7a466; 

V      64/3  4/1' 

7.    Va^yy  =  aj2#83  ; 


v- 


ltKi^io  _  — ary 


107.  Square  Root  of  a  Polynomial.  —  Since  the 
square  of  a  +  6  is  a2  -h  2a6  -f  62,  the  square  root  of 
cr  +  2ab  -+-  b2  is  a  +  b.  We  may  deduce  the  general  rule 
for  the  extraction  of  the  square  root  of  a  polynomial  by 
observing  in  what  manner  a  +  b  may  be  derived  from 
a2  -f  2ab  +  &2. 

Arrange  the  terms  of  the  square  according  to  the  descend- 
ing powers   of   a ;    then  the   first 
term  is  a2,  and  its  square  root  is        ft2  +  2ab  +  &2  |fl  -f-  & 

a,  which  is  the  first  term  of  the       ^ 

required  root.     Subtract  its  square,    2a  -f-  b)2ab  -f-  b2 
a2,  from  the  given  expression,  and  2ab  +  b2 

bring  down  the  remainder,  2ab  +  ft2. 

Thus,  6,  the  second  term  of  the  root,  will  be  the  quotient 
when  2a&,  the  first  term  of  the  remainder,  is  divided  by  2a, 
i.e.,  by  double  the  first  term  of  the  root.  This  second  term, 
6,  added  to  2a,  twice  the  first  term,  completes  the  divisor, 
2a  +  b ;  multiply  this  complete  divisor  by  6,  the  second 
term,  and  subtract  the  product,  i.e.,  2ab  +  b2,  from  the 
remainder,  and  the  operation  is  completed. 

If  there  were  more  terms  we  should  proceed  with  a  -f  b 
as  we  have  done  with  a ;  its  square,  a2  -f-  2ab  +  &2,  has 
already  been  subtracted  from  the  given  expression,  so  we 
should  divide  the  remainder  by  twice  the  first  term,  i.e.,  by 
2(a  4-  &),  for  a  new  term  of  the  root.  Then  for  a  new 
subtrahend  we  should  multiply  the  sum  of  2  (a  +  b)  and  the 


206  EXAMPLES. 

new  term  by  the  new  term.     The  process  must  be  continued 
till  the  required  root  is  found. 

Hence  to  extract  the  square  root  of  a  polynomial,  we  have 
the  following 

Rule. 

Arrange  the  terms  according  to  the  powers  of  some  letter; 
find  the  square  root  of  the  first  term  for  the  first  term  of  the 
square  root;  place  this  on  the  right,  and  subtract  its  square 
from  the  given  polynomial. 

Double  the  root  already  found  for  a  tried  divisor;  divide 
the  first  term  of  the  remainder  by  this  trial  divisor  for  the 
second  term  of  the  root,  and  annex  this  second  term  to  the  root 
and  also  to  the  trial  divisor  for  the  complete  divisor. 

Multiply  the  complete  divisor  by  the  second  term  of  the  root, 
and  subtract  the  product  from  the  remainder. 

If  there  are  other  terms  remaining,  repeat  the  process  until 
there  is  no  remainder,  or  until  all  the  terms  of  the  root  have 
been  obtained. 

EXAMPLES. 

1.    Find  the  square  root  of 

4a;4  -  20a;3  +  37a;2  -  30a;  +  9  |2a;2  -  5x  -f-  3. 


4a; 


4a;2  —  5a; 


-  20a;3  +  37a;2 

-  20a-3  +  25a;2 


4a;2  -  10a;  4  3 


12a;2  -  30a;  +  9 
12a;2  -  30s  4  9 


The  expression  is  arranged  according  to  the  descending 
powers  of  x. 

The  square  root  of  4a;4  is  2a;2,  and  this  is  placed  at  the 
right  of  the  given  expression  for  the  first  term  of  the  root. 
By  doubling  this  we  obtain  4  a;2,  which  is  the  trial  divisor. 
The  second  term  of  the  root,  —ox,  is  obtained  by  dividing 
—  20a-3,  the  first  term  of  the  remainder,  by  •l.r2.  and  this  new 
term  has  to  be  annexed  both  to  the  root  and  divisor.     Next 


EXAMPLES.  207 

multiply    the   complete    divisor   by    —  5x   and   subtract   the 
product  from  the  first  remainder. 

We  then  double  the  root  already  found  and  obtain 
Ax2  —  10a  for  a  new  trial  divisor.  Dividing  12a2,  the  first 
term  of  the  remainder,  by  4a;'2,  the  first  term  of  the  divisor, 
we  get  3,  which  we  annex  both  to  the  root  and  divisor.  We 
now  multiply  the  complete  divisor  by  3  and  subtract.  There 
is  no  remainder,  and  the  root  is  found. 

2.    Fiud  the  square  root  of 
15aV  -  6aa5  +  a6  -  20a3a3  +  a6  +  15a4a2  -  Ga5x. 

Arrange  in  descending  powers  of  x. 
jB6-6ajB5+15a3aJ*-20aV+  15a4a2-6a5a+a6  \x3-3ax2+3a2x-a* 


2x3-3ax2 


-6aa5+15a2a4 
■6ax5+  9a2a4 


2x?-6ax2+3a2x 


6aV-20aV+15a4aj2 

6a2a4-18a3a3+  9a4a2 


2xs-6ax2+6a2x-a3 


2a3x3+  SaW-Stfx+a* 
2a3x3+  6a4a2-6a5a+a6 


Note.  —  All  even  roots  admit  of  a  double  sign  (Art.  106).  Thus 
the  square  root  of  a2  +  lab  +  b2  is  either  a  +  b  or  —  a  —  b,  as 
may  be  verified.  In  the  process  of  extracting  the  square  root  of 
a2  +  2ab  -\-  b2,  we  begin  by  extracting  the  square  root  of  «2,  and  this 
may  be  either  a  or  —a.  If  we  take  the  latter,  and  continue  the 
operation  by  the  rule  as  before,  we  shall  obtain  —  a  —  b.  A  similar 
remark  holds  in  every  other  case.  Thus,  in  Ex.  2  the  square  root  of 
the  first  term  x6  is  either  x3  or  — x3.  If  we  take  the  latter,  and  continue 
the  operation  as  before,  we  shall  obtain  —  x3  -f-  3ax2  —  3a2x  -f  a3. 

Since  the  fourth  power  is  the  square  of  the  square,  the 
fourth  root  of  an  expression  may  be  found  by  extracting 
the  square  root  of  the  square  root.  Similarly  the  eighth  root 
may  be  found  by  three  successive  extractions  of  the  square 
root ;  and  so  on. 

For  example,  required  the  fourth  root  of 

16a4  -  96afy  +  216x2y2  -  21Gav/3  +  61y\ 


208 


EXAMPLES. 


By  the  rule  we  find  that  the  square  root  is  4a,*2  —  12xy  -f-  9y2 ; 
and  the  square  root  of  this  is  2x  —  3?/,  which  is  therefore 
the  fourth  root  of  the  given  expression. 

3.    Find  the  square  root  of 


16y 


Sx 


:r 


24  +  ^f  -  -  +  -  -  ^ 


32?/ 
x*         y       y        x 
Arranging  in  descending  powers  of  ?/,  we  have 


16.v2       32?/ 


Sx 


_  ^i  +  24  -  —  +  -a 


*y  -  4  +  • 


x 


-  4 

_  3J/  +  24 
-  ^  +  16 

X 

8 

i 

b               y 

8  - 

_  Sx       x2 

y      yl 

8  - 

_  Sx       x2 

y      y* 

Here  the  second  term  of  the  root,  —4,  is  found  by  the 

rule  as  usual,  i.e.,  by  dividing  —  °—  by   — ,  and  the  third 

x  x 

x  Sv 

term,  -,  is  found  by  dividing  8  by  — '. 

y  * 

EXAMPLES. 

Find  the  square  root  of  each  of  the  following : 

4.  2bx2  —  dOxy  +  9#2.  Ans.  5x  —  3y. 

5.  4z4  -  12s8  +  29x2  -  30a;  +  25.  2x2  -  Sx  +  5. 

6.  1  _  10aj  +  27x2  -  lO.r3  +  x4.  1  -  5x  +  x2. 

7.  4z2-f-  9#2+  25z2+  1 2a#  -  dOyz  -  20a».      2jc  -f  3?/  -  .r>z. 

8.  34&8-22<B*+aJ8+  121a2- 374a?  +  289.     a-3-  llaj+17. 


*  The  reason  for  this  arrangement  will  appear  in  Chap.  XII. 


SQUARE   ROOT   OF  ARITHMETIC  NUMBERS.  209 

n     64SB2    .    32.x*    .    .  A        8a;    .    n 

9. -\ h  4.  Arts. h  2. 

%*         3^  3z/ 

10.    —  +  -  H aa;  +  —  -  2.  -  H . 

4         cr       a;  ar  2         a*       a 

108.  Square  Root  of  Arithmetic  Numbers.  —  The 

rule  which  is  given  in  Arithmetic  for  extracting  the  square 
root  of  a  number  is  based  upon  the  method  explained  in  Art. 
107. 

Since  1  =  l2,  100  =  102,  10000  =  1002,  1000000  =  10002,- 
and  so  on,  it  follows  that  the  square  root  of  a  number 
between  1  and  100  is  between  1  and  10;  the  square  root  of 
a  number  between  100  and  10000  is  between  10  and  100 ; 
the  square  root  of  a  number  between  10000  and  1000000 
is  between  100  and  1000,  and  so  on.  That  is,  the  square 
root  of  a  number  of  one  or  two  figures  consists  of  only  one 
figure  ;  the  square  root  of  a  number  of  three  or  four  figures 
consists  of  two  figures ;  the  square  rout  of  a  number  of  five 
or  six  figures  consists  of  three  figures ;  and  so  on.  Hence 
the 

Rule. 

If  a  point  is  placed  over  every  second  figure  in  any  number, 
beginning  with  the  units'  place,  the  number  of  points  ivill  show 
the  number  of  figures  in  the  square  root. 

Find  the  square  root  of  5329. 

Point  the  number  according  to  the  rule.  Thus,  it  appears 
that  the  root  consists  of  two  places  of  figures,  i.e.,  of  tens 
and  units.     Let  a  denote  the 

value  of  the  figure  in  the  tens'  5329(70  +  3  =  73. 

place  of  the  root,  and  b  that  4900 


429 
429 


in  the  units'  place.     Then  a 

must  be  the  greatest  multiple     140  -f  3 

of  10  whose  square  is  less  than 

5300;  this  we  find  to  be  70.     Subtract  a2,  i.e.,  the  square 

of  70,  from  the  given  number,  and  the  remainder  is  429, 

which  must  equal  {2a  -f-  b)b.     Divide  this  remainder  by  the 


210  SQUARE  ROOT   OF  ARITHMETIC  NUMBERS. 

trial  divisor,  2a,  i.e.,  by  140,  and  the  quotient  is  3,  which 
is  the  value  of  b.  Then  the  complete  divisor,  2a  +  b,  is 
140  +  3  =  143,  and  (2a  4-  o)b,  that  is,  143  x  3  or  429 
is  the  number  to  be  subtracted ;  and  as  there  is  now  no 
remainder,  we  conclude  that  70  -f  3  or  73  is  the  required 
square  root. 

In  squaring  the  tens,  and  also  in  doubling  them,  the 
ciphers  are  omitted  for  the  sake  of  brevity,  though  they  are 
understood.  Also  the  units'  figure  is  added  to  the  double  of 
the  tens  by  merely  writing  it  in  the  units'  place.  The  actual 
operation  is  usually  performed  as  follows : 

If  the  root  consists  of  three  places  of  figures,  let  a  repre- 
sent the  hundreds  and  b  the  tens ;  then  hav- 
ing obtained  a  and  b  as  before,  let  a  represent  5329(73 
the  hundreds  and  tens  as  a  new  value ;  and            49 
find  a  new  value  of  b  for  the  units  ;  and  in    1 43  \  429 
general,  let  a  represent  the  part  of  the  root  429 
already  found. 

Hence  for  the  extraction  of  the  square  root  of  a  number, 
we  have  the  following 

Rule. 

Separate  the  given  number  into  periods  of  two  figures  each, 
by  pointing  every  second  figure,  beginning  at  the  units'  place. 

Find  the  greatest  number  ivhose  square  is  contained  in  the 
left  period,  and  place  it  on  the  right;  this  is  the  first  figure 
of  the  root  ;  subtract  its  square  from  the  first  period,  and  to 
the  remainder  bring  down  the  next  period  for  a  dividend. 

Double  the  root  already  found  for  a  trial  divisor,  and  see 
how  many  times  it  is  contained  in  the  dividend,  omitting 
the  last  figure,  and  annex  the  result  to  the  root  and  cdso  to  the 
tibial  divisor. 

Multiply  the  divisor  thus  completed  by  the  figure  of  the  root 
last  obtained,  and  subtract  the  product  from  the  dividend. 

If  there  are  more  periods  to  be  brought  down,  continue  the 
operation  as  before,  regarding  the  root  already  found  as  one 
term. 


SQUARE  ROOT  OF  A   DECIMAL.  211 

Extract  the  square  root  of  132496,  and  10246401. 
1.    132496(364  2.    1024640i(3201 

9  9 

66)424  62)124 

396  124 


724)2896  6401)6401 

2896  6401 


As  the  trial  divisor  is  an  incomplete  divisor,  it  is  sometimes 
found  that  the  product  of  the  complete  divisor  by  the  corre- 
sponding figure  of  the  root  exceeds  the  dividend.  In  such 
a  case  the  last  root  figure  must  be  diminished.  Thus,  in 
Ex.  1,  after  finding  the  first  figure  of  the  root,  we  are  re- 
quired by  the  rule  to  divide  42  by  6  for  the  next  figure  of 
the  root,  so  that  apparently  7  is  the  next  figure.  On  multi- 
plying however  67  by  7  we  obtain  a  product  which  is  greater 
than  the  dividend  424,  which  shows  that  7  is  too  large,  and 
we  accordingly  try  6,  which  is  found  to  be  correct. 

The  student  will  observe  in  Ex.  2  that,  in  consequence  of 
the  dividend,  exclusive  of  the  right  hand  figure,  not  contain- 
ing the  trial  divisor,  64,  we  place  a  cipher  in  the  root  and 
also  at  the  right  of  the  trial  divisor  64,  making  it  640 ;  we 
then  bring  down  the  next  period  and  proceed  as  before. 

109.  Square  Root  of  a  Decimal.  —  The  rule  for 
extracting  the  square  root  of  a  decimal  follows  from  the  rule 
of  Art.  108.  If  any  decimal  be  squared  there  will  be  an 
even  number  of  decimal  places  in  the  result;  thus  (.25)2  = 
.0625,  and  (.111)2  =  .012321.  Therefore  there  cannot  be 
an  exact  square  root  of  any  decimal  which  has  an  odd 
number  of  decimal  places. 

The  square  root  of  32.49  is  one-tenth  of  the  square  root 
of  3249.  Also  the  square  root  of  .0361  is  one-hundredth  of 
that  of  361.  Hence,  for  the  extraction  of  the  square  root 
of  a  decimal,  we  have  the  following 


212  SQUARE  ROOT  OF  A  DECIMAL. 

Rule. 

Separate  the  given  number  into  periods  of  two  figures  each, 
by  putting  a  point  over  every  second  figure,  beginning  at  the 
units'  place  and  continuing  both  to  the  right  and  to  the  left  of 
it;  then  proceed  as  in  the  extraction  of  the  square  root  of  in- 
tegers, and  point  off  as  many  decimal  places  in  the  result  as 
there  are  periods  in  the  decimal  part  of  the  proposed  number. 

If  there  be  a  final  remainder  in  extracting  the  square  root 
of  an  integer,  it  indicates  that  the  given  number  has  not  an 
exact  square  root.  We  may  in  this  case  place  a  decimal 
point  at  the  end  of  the  given  number,  and  annex  any  even 
number  of  ciphers,  and  continue  the  operation  to  any  desired 
extent.  We  thus  obtain  a  decimal  part  to  be  added  to  the 
integral  part  already  found. 

Also,  if  a  decimal  number  has  no  exact  square  root,  we 
may  annex  ciphers  and  obtain  decimal  figures  in  the  root  to 
any  desired  extent.  ^_  ^ 

Find  the  square  root  oij  12  ;  and  also  or  .4  to  three  deci- 
mal places. 

12.00o6o6(3.464  .400006(.632 

9  36 


64)300  123)400 

256  369 


686)4400  1262)3100 

4116  2524 


6924)28400 
27696 


Note.  —  We  see  here  in  what  sense  we  can  be  said  to  approximate 
to  the  square  root  of  a  number.  The  square  of  3.464  is  less  than  12, 
and  the  square  of  3.405  is  greater  than  12.  Also  the  square  of  .032  is 
less  than  .4,  and  the  square  of  .033  is  greater  than  .4. 

No  fraction  can  have  a  square  root  unless  the  numerator  and 
denominator  are  both  square  numbers  when  the  fraction  is  in  its  low- 
est terms.  But  we  may  approximate  to  the  square  root  of  a  fraction 
to  any  desired  extent.     Thus, 


SQUARE  ROOT   OF  A    DECIMAL.  213 

Let  it  be  required  to  find  the  square  root  of  ^. 
Here  (Art.  Ill)  y/|  =  j| 

Therefore  we  find  the  square  root  of  5   and  also  of  7,  ap- 
proximately, and  divide  the  former  by  the  latter. 

Or,  we  may  reduce  the  fraction  |  to  a  decimal  to  any 
required  degree  of  approximation,  and  obtain  the  square 
root  of  this  decimal. 

Otherwise  thus : 

X^7       \^5  x  7       V^35. 


vf-v/i 


X  7       v^7  x  7  7 

then  find  the  square  root  of  35  approximately,  and  divide 
the  result  by  7.  Either  of  these  last  methods  is  preferable 
to  the  first. 

//  the  square  root  of  a  number  consists  of  2n  +  1  figures,  when 
the  first  n  +  1  of  these  have  been  obtained  by  the  ordinary  method,  the 
remaining  n  may  be  obtained  by  division. 

Let  N  represent  the  given  number ;  a  the  part  of  the  square  root 
already  found,  i.e.,  the  first  n  +  1  figures  found  by  the  rule,  with  n 
ciphers  annexed;  and  x  the  part  of  the  root  which  remains  to  be 
found. 

Then  yfy  =  a  +  x\ 

N  =  a2  +  2ax  +  x2\ 

...     *^  =  x  +  t (1) 

2a  2a 

Now  N  —  a2  is  the  remainder  after  n  +  1  figures  of  the  root,  rep- 
resented by  a,  have  been  found;  and  2a  is  the  corresponding  trial 
divisor.     We  see  from  (1)  that  N  —  a2  divided  by  2a  gives  x,  the  rest 

x2 
of  the  square  root  required,  increased  by  — 

2a 

x2 

Now  —  is  a  proper  fraction,  so  that  by  neglecting  the  remainder 

2a 

arising  from  the  division,  we  obtain  x,  the  rest  of  the  root.     For,  x 

contains  n  figures  by  supposition,  so  that  x2  cannot  contain  more  than 

2n  figures;  but  a  contains  2n  +  1  figures  (the  last  n  of  which  are  ciphers) 

r,2 
and  thus  2a  contains  2n  +  1  figures  at  least;  therefore  —  is  a  proper 

2a 
fraction. 

From  this  investigation,  by  putting  n  =  1,  we  see  that  at  least  two 
of  the  figures  of  a  square  root  must  have  been  obtained  in  order  that 


214  CUBE  ROOT  OF  A  POLYNOMIAL. 

the  method  of  division  may  give  the  next  figure  of  the  square  root 
correctly. 

We  will  apply  this  method  to  finding  the  square  root  of  290  to  five 
places  of  decimals.  We  must  obtain  the  first  four  figures  in  the  square 
root  by  the  ordinary  method ;  and  then  the  remaining  three  may  be 
found  by  division. 

290  (17.02 

1 

27)  190 
189 
3402)  10000 
6804 
3196 
We  now  divide  the  remainder  3196,  which  is  N  —  a2,  by  twice  the 
root  already  found,  3404,  which  is  2a,  and  obtain   the  next  three 
figures.    Thus, 

3404)  31960  (938 
30636 
13240 
10212 


30280 
27232 


3048 
Therefore  to  five  places  of  decimals,  y/290  =  17.02938. 
In  extracting  the  square  root,  the  student  will  observe  that  each 
remainder  brought  down  is  the  given  expression  minus  the  square  of 
the  root  already  found,  and  is  therefore  in  the  form  N  —  a2. 

EXAMPLES. 

Find  the  square  roots  of  the  following  numbers : 


1. 

15129. 

Ans.  123. 

5. 

.835396. 

Ans.   .914. 

2. 

103041. 

321. 

G. 

1522756.. 

1234. 

3. 

3080.25. 

55.5. 

7. 

29376400. 

5420. 

4. 

41.2164. 

6.42. 

8. 

364524.01. 

620.1. 

110.  Cube  Root  of  a  Polynomial.  —  Since  the  cube 
of  a  +  b  is  a3  -f  Sa2b  +  oab'1  -f-  b3,  the  cube  root  of 
a3  +  Sa2b  +  3ab2  +  bs  is  a  +  b.  We  may  deduce  a 
general  rule  for  the  extraction  of  the  cube  root  of  a  poly- 


CUBE   ROOT   OF  A    POLYNOMIAL.  215 

noraial  by  observing  in  what  manner  a  +  b  may  be  derived 
from  a3  +  3d2b  +  3ab2  +  b\ 

Arrange  the  terms  of  the  cube  according  to  the  descending 
powers  of  a ',  then  the  first  term  is  a3,  and  its  cube  root  is  a, 
which  is  the  first  term  of  the  required  root.  Subtract  its 
cube,  a3,  from  the  given  expression,  and  bring  down  the 
remainder  Sa2b  -f-  3ab2  -b  63.  Thus,  6,  the  second  term  of 
the  root,  will  be  the  quotient  when  3a26,  the  first  term  of  the 
remainder,  is  divided  by  3a2,  i.e.,  by  three  times  the  square 
of  the  first  term  of  the  root. 

Also,  since  3a2b  -f  Sab2  -f  63  =  (3a2  +  3a5  +  b2)h,  we 
add  to  the  trial  divisor  Sab  -f  62,  i.e.,  three  times  the 
product  of  the  first  term  of  the  root  by  the  second,  plus 
the  square  of  the  second,  and  we  have  the  complete  divisor 
3a2  +  3ab  -f-  b2 ;  multiply  this  complete  divisor  by  6,  and 
subtract  the  product,  3a26  -f-  3ab2  +  63,  from  the  remainder, 
and  the  operation  is  completed. 

The  work  may  be  arranged  as  follows : 

a3  +  3crb  +  3ab2  4-  bs  [a  +  b 
a3 


3a2  +  Sab  +  b- 


Serb  -f  3a62  +  & 
Sa2b  -f  Sab2  +  &1 


If  there  were  more  terms,  we  should  proceed  with  a  -f-  b 
as  we  have  done  with  a  ;  its  cube,  a3  +  3a26  +  Sab2  +  o3, 
has  already  been  subtracted  from  the  given  expression,  so 
we  should  divide  the  remainder  by  three  times  the  square  of 
the  first  term,  i.e.,  by  3 (a  +  6)2,  for  a  new  term  of  the  root, 
c  say.  Then  for  a  new  complete  divisor  we  would  have 
3  (a  +  6)2  +  3  (a  +  b)c  -f-  c2 ;  and  this  multiplied  by  c 
would  give  us  a  new  subtrahend  ;  and  so  on.  Hence  the 
following 

Rule. 

Arrange  the  terms  according  to  the  poicers  of  some  letter; 
find  the  cube  root  of  the  first  term  for  the  first  term  of  the 
cube  root;  and  subtract  its  cube  from  the  given  polynomial. 


216 


EXAMPLES. 


Take  three  times  the  square  of  the  root  already  found  for  a 
trial  divisor ;  divide  the  first  term  of  the  remainder  by  this 
trial  divisor  for  the  second  term  of  the  root ;  annex  this  second 
term  to  the  root,  and  complete  the  divisor  by  adding  to  the  tried 
divisor  three  times  the  product  of  the  first  and  second  terms 
of  the  root  and  the  square  of  the  second  term. 

Multiply  the  complete  divisor  by  the  second  term  of  the  rooty 
and  subtract  the  product  from  the  remainder. 

If  there  are  other  terms  remaining,  take  three  times  the 
square  of  the  part  of  the  root  already  found  for  a  new  trial 
divisor ;  and  continue  the  operation  until  there  is  no  remain- 
der, or  until  all  the  terms  of  the  root  have  been  obtained. 


EXAMPLES. 

1.    Find  the  cube  root  of  8a;3  —  S6x2y  +  54a;?/2  —  27if. 
The  work  may  be  arranged  as  follows : 

8x3-36x2y+54;xy2-27y3\2x  -  Sy 

8a;3 


3  (2a)2  =12a2 

S(2x)(-3y)  =        -IS*?/ 

(-3y)2=  +v 


12x2-18xy+dyc< 


2.    Find  the  cube  root  of 


-3Gx2y+5±xy2-27y3 
-36x2y+54xy2-27y3 


|3+4z-2.r2 
27  +  108;s+  90a;2-  80a*3- G0a;4-f- 48a;5 -8a;c 
27 


27+36a;+16a;2 

27  +  72a+48a;2 

-18a;2 -24a-3 


108a;-f-  90a;2-  80a;a 
108a;+144a;2+   G4a;a 


+4a*4 


27  +  72a«+30a;2-24a;3+la;4 


54a;2- 144a;3- C0a-4-f48a-5-8a;e 


_r)Jar2-144a-3-60.r4+lS.rr,-8a;6 


Explanation.  — The  root  is  placed  above  the  given  expression  for 
convenienee.     When  we  have  obtained  two  terms  in  the  root,  "J  +  4x, 


CUBE  ROOT  OF  ARITHMETIC  NUMBERS,  217 

we  form  the  second  divisor  as  follows:  take  3  times  the  square  of  the 
root  already  found  for  the  trial  divisor,  27  +  72x  +  48a;2;  divide  —54a;2, 
the  first  term  of  the  remainder,  by  27,  the  first  term  of  the  trial 
divisor;  this  gives  the  third  term  of  the  root,  — 2.r2.  To  complete  the 
divisor  we  add  to  the  trial  divisor  3  times  the  product  of  (3  +  4./ )  and 
—2x2,  and  also  the  square  of  —  2x~.  Xow  multiply  the  complete  divisor 
by  — 2x2  and  subtract;  there  is  no  remainder  and  the  root  is  found. 

Find  the  cube  root  of  each  of  the  following : 

3.  a9  4-  3d2  4-  3«  4-  1.  Ans.  a  +  1. 

4.  64a3  -  Uicrb  +  lOSab2  -  276s.  4a  -  36. 

5.  x*  4-  Sx5  4-  6x4  4-  7x3  4-  Oar  +  Sx  +  1.  x*  +  x  +  1. 

6.  l-6x+21x2-Uxs+63xi-5±x5  +  27xQ.  l-2x+3x2. 

111.  Cube  Root  of  Arithmetic  Numbers. — The 

rule  which  is  given  in  Arithmetic  for  extracting  the  cube  root 
of  a  number  is  based  upon  the  method  explained  in  Art.  110. 

Since  1  =  l3,  1000  =  103,  1000000  =  1003,  and  so  on,  it 
follows  that  the  cube  root  of  a  number  between  1  and  1000 
is  between  1  and  10;  the  cube  root  of  a  number  between 
1000  and  1000000  is  between  10  and  100  ;  and  so  on.  That 
is,  the  cube  root  of  a  number  of  one  or  two  or  three  figures 
consists  of  only  one  figure  ;  the  cube  root  of  a  number  of 
four  or  jive  or  six  figures  consists  of  two  figures  ;  and  so  on. 
Hence  the 

Rule.  If  a  point  is  placed  over  every  third  figure  in  any 
number,  beginning  with  the  units7  place,  the  number  of  points 
will  show  the  number  of  figures  in  the  cube  root. 

Find  the  cube  root  of  614125. 

Point  the  number  according  to  the  rule.  Thus  it  appears 
that  the  root  consists  of  two  places  of  figures,  i.e.,  of  tens 
and  units.  Let  a  denote  the  value  of  the  figure  in  the  tens' 
place  of  the  root,  and  b  that  in  the  units'  place.  Then  a 
must  be  the  greatest  multiple  of  10  whose  cube  is  less  than 
614000  ;  this  we  find  to  be  80.  Subtract  a3,  i.e.,  the  cube 
of  80,  from  the  given  number,  and  the  remainder  is  102125, 
which  must  equal  (3a2  4-  oab  4-  b'2)b.  Divide  this  remainder 
by  the  trial  divisor,  3a'2,  i.e.,  by  19200,  and  the  quotient  is  5, 


218  CUBE  ROOT   OF  ARITHMETIC  NUMBERS. 

which  is  the  value  of  b.  Then  adding  Sab,  or  1200,  and  b2, 
or  25,  to  the  trial  divisor  3a2,  or  19200,  we  obtain  the  com- 
plete divisor  20425  ;  and  multiplying  the  complete  divisor 
by  5  and  subtracting  the  product  102125,  there  is  no  remain- 
der.    Therefore  85  is  the  required  cube  root. 

614125  1 80  -f  5 
512000 


3a2  =  3(80)2       =  19200 
Sab  =  3(80)  (5)  =     1200 

b2  =  (5)2  = 25 

20425 


102125 


102125 


In  cubing  the  tens  the  ciphers  are  omitted  for  the  sake  of 
brevity,  though  they  are  understood. 

If  the  root  consists  of  three  places  of  figures,  let  a  repre- 
sent the  hundreds  and  b  the  tens,  and  proceed  as  before. 
See  Art.  108. 

Hence  for  the  extraction  of  the  cube  root  of  a  number, 
we  have  the  following 

Rule. 

Separate  the  given  number  into  periods  of  three  figures 
each  by  pointing  every  third  figure,  beginning  at  the  units' 
place. 

Find  the  greatest  number  whose  cube  is  contained  in  the  left 
period,  and  place  it  on  the  right;  this  is  the  first  figure  of  the 
root;  subtract  its  cube  from  the  first  period,  and  to  the  re- 
mainder bring  down  the  next  period  for  a  dividend. 

Take  three  times  the  square  of  the  root  already  found  for 
a  trial  divisor,  and  see  hoio  many  times  it  is  contained  in  the 
dividend,  omitting  the  last  tivo  figures,  and  annex  the  result 
to  the  root.  Add  together,  the  trial  divisor  with  two  ciphers 
annexed;  three  times  the  product  of  the  last  figure  of  the  root 
by  the  rest,  ivith  one  cipher  annexed ;  and  the  square  of  the 
last  figure  of  the  root. 


CUBE   ROOT   OF  A    DECIMAL.  219 

Multiply  the  divisor  thus  completed  by  the  figure  of  the  root 
last  obtained,  and  subtract  the  product  from  the  dividend. 

If  there  are  more  peiiods  to  be  brought  down,  the  operation 
must  be  repeated,  regarding  the  root  already  found  as  one  term. 

Extract  the  cube  root  of  109215352. 


109215352(478 

64 

3(4)2  =  4S00 

45215 

3(4) (7)  =  840 

(7)2  =   49 

5689 

39823 

3(47)2  =  662700 

5392352 

3(47)  (<s)  =  11280 

(8)2  =     64 

674044 

5392352 

After  finding  the  first  figure  of  the  root  we  are  required  by 
the  rule  to  divide  452  by  48  for  the  next  figure  of  the  root, 
so  that  apparently  8  or  9  is  the  next  figure.  On  trial  we  find 
that  these  numbers  are  too  large  ;  so  we  try  7,  which  is  found 
to  be  correct.  As  in  the  case  of  the  square  root  (Art.  113), 
we  are  liable  occasionally  to  try  too  large  a  figure,  especially 
at  the  early  stages  of  the  operation. 

112.  Cube  Root  of  a  Decimal.  —  If  the  cube  root 
have  any  number  of  decimal  places,  the  cube  will  have  three 
times  as  many.  Hence  for  the  extraction  of  the  cube  root 
of  a  decimal,  we  have  the  following 

Rule. 

Separate  the  given  number  into  periods  of  three  figures 
each,  by  'putting  a  point  over  every  third  figure,  beginning  at 
the  units'  place  and  continuing  both  to  the  right  and  to  the  left 
of  it;  then  proceed  as  in  the  extraction  of  the  cube  root  of 
integers,  and  point  off  as  many  decimal  jylaces  in  the  result 
as  there  are  periods  in  the  decimal  part  of  the  proposed 
number. 


220 


EXAMPLES. 


If  there  be  a  final  remainder  in  extracting  the  cube  root 
of  any  number,  integral  or  decimal,  it  indicates  that  the 
number  has  no  exact  cube  root.  We  ma}'  in  this  case,  as  in 
the  extraction  of  the  square  root  (Art.  109),  annex  any 
number  of  ciphers,  and  continue  the  operation  to  any  desired 
extent. 

Extract  the  cube  root  of  1481.544 


1481.544 | 11.4 

1 

300 

481 

30 

1 

331 

331 

36300 

150544 

1320 

16 

37636 

150544 

The  cube  root  is  11.4. 

EXAMPLES. 


Show  that 

1. 

(- 

-2a5)4  = 

16a2" 

2. 

(- 

-a4)5  =  - 

-a20. 

3. 

(- 

-3a765c)3 

=  — 

27a2VV3. 

4. 

(" 

-5aW)3 

=  — 

125a669c12. 

5.  (-3a7£/2c)4  =  81a2W. 
Find  the  value  of 

6.  (-ax'  +  by*)2. 

7.  (2a4  +  363)2. 

8.  (2a2  -  3U2)3. 

9.  (2x  +  3)4. 
10.  (1  +  x)*  -  ( 


Arts.  a2xB  -  2abxY  +  b2y\ 

4a8  +  12a4^3  +  966. 

8a<>  _  QOaVr  +  Mcrb*  -  276°. 

16.r4  +  !)6.t:3  +  2163s  +  216a  +  81. 

-  a?)5.  2(5ae  +  lO.f3  +  x5). 


EXAMPLES. 


221 


11.  (1  —  &t  +  3a;2)2.  1  -  6a:  +  15a;2  -  18a"  +  9a4. 

12.  (2  +  3.7j+4.r2)2+(2-3.i-+4a;2)2.       2(4  +  25ar+16a;4). 

13.  (1  +  3x  +  2a;2)3. 

^4?*s.  1  +  9a;  +  33ar  +  63a;3  +  GGo;4  -f  36a5  +  8a;6. 

14.  (2  +  3a-  +  4a;2)3  -  (2  -  3s  +  4a;2)3. 

4«s.  2(36a;  +  171a,-3  +  144a,-5). 

15.  (1  +  4a!  +  6x2  +  4xs  -f  a;4)2. 

>/is.  l+8a;+28a;2+56x'3+70aj4+5Ga5+28a;6+8a;7+a;8. 

Show  that 

18.    y/-32a;iy5  =  -2x2y\ 


16.    y/32xPyw  =  2a;?/2 


17.    y/256af 


19. 


V 


eg 


=  2aar.  V  527^ 

Find  the  square  roots  of  the  following  expressions  : 

20.  9a-4  -  12or3  -  2a-2  +  4a;  +  1.       Ans.  3ar  -  2x  -  1, 

21.  16a;6  +  16a;7  -  4a;8  -  4a-9  -f  a;10.         4a;3  +  2a;4*-  a;5, 


22.  25a?4— 30aa*+49aV— 24a8a;+16a4. 

23.  x*  -  4a3  +  Sx  +  4. 

24.  x*+  lax5—  lOaV-f  4a5a;  +  a6.       a;3 


oar 


3aa;+4a2. 

0;2  _   2.7;  _   2. 

2«a*2—  2a?x—as. 


25.  a;4  -  2«a;3  +  (a2  +  2b2)x2  -  2db*x  +  b\       x2  -  «a;  +  b2. 

26.  16  -  96a;  +  216a?  -  216a-3  +  81a-4.       4  -  12a;  +  9.r. 

27.  9a-6-12a;5-f  22a-4+ar+12a;  +  4.       3a-3-  2a:2  +  3a-  +  2. 


28.  4a;8  -  4a;6  -  7a-4  +  4a;2  +  4. 

29.  1  —  xy  —  ^x2i/2  +  2a-y  +  4«y. 


2a:' 


a- 


2. 


30. 


+  4x-2  -f 


4  3 

Find  the  square  roots  of 

31.  165649.  Ans.  407. 

32.  384524.01.  621.1. 

33.  4981.5364.  70.58. 

34.  .24373969.  .4937. 

35.  144168049.  12007. 


?,-3 


36 


4  cue 

3  ' 


1  -  \xy  —  2a-y2. 

x-  -  2x  +  C'- 
2  3 


.5687573056.  Ans.  .75416. 

37.  3.25513764.  1.8042. 

38.  4.54499761.  2.1319. 

39.  196540602241.      443329. 


52                                             EXAMPLES. 

Find  the  square  roots  of  the  following  to  five  decimals : 

40.    .9.             Ans.  .94868. 

43.    .00852.         Ans.  .09230 

41.    6.21.                2.49198. 

44.    17.                         4.12310 

42.    .43.                     .65574. 

45.    129.                    11.35781 

Find  the  cube  roots  of  the  following  expressions : 

46.  1728a;6  +  1728a?4?/3  +  576a2?/6  +  64?/9.     Ans.  12a;2 +  4y\ 

47.  a;6  —  Sax5  -\-  ba3x3  —  3a5x  —  a6.  x2  —  ax  —  a2. 

48.  8a;6  +  48ca;5  +  60c2a;4  -  80csxs  -  90c4a;2  -f-  108c5a;  -  27c6. 

Ans.  2x2  +  icx  —  3c2. 

49.  i  _  9a  +  39^2  _  99^3  +  15Qsa  _  144aji  +  64a.6> 

Ans.  1  —  3a;  -J-  4a;2. 

50.  27a;6  -  27a,-5  -  18a;4  +  17a;3  +  6a;2  -  3x  -  1. 

Ans.  3x2  —  x  —  1. 


51 


x 


,8  _i_  &£  +  ^  _  ?/  +  %!  _  ^ 
?/3        ?/2         ?/        a;3        a;2         a; 


4. 


2  +  2  -  -V 

2/  « 


Find  the  fourth  roots  of 

52.  1  +  4a;  -f-  Gx2  -f  4a;3  +  x\  Ans.  1  +  x. 

53.  1  -  4a;  +  10a;2  -  16a;3  +  19a;4  -  16a;5  +  10a;6  -  4a;7  +  x\ 

Ans.  1  —  x  +  a;2. 
Find  the  sixth  root  of 

54.  1  +  12a;  +  60a;2  +  160a;3  +  240a;4  +  192a;5  +  64a;6. 

Ans.  1  -f-  2x. 
Find  the  cube  roots  of 


55.  2628072.      Ans.  138. 

56.  3241792.  148. 

57.  60236.288.  39.2. 


58.  .220348864.  Ans.  .604. 

59.  1371330631.  1111. 

60.  20910518875.        .2755. 


THE   THEORY  OF  EXPONENTS  —  SURDS.  223 


CHAPTER     XII. 

THE  THEORY  OF  EXPONENTS  — SURDS. 

113.  Exponents    that    are    Positive    Integers.  — 

Hitherto  we  have  supposed  that  an  exponent  was  always  a 
])ositive  integer.      Thus,  in  Art.  12,  we   defined   am  as  the 
product  of  m  factors  each  equal  to  a,  which  would  have  no 
meaning  unless  the  exponent  was  a  positive  integer. 
When  m  and  n  are  positive  integers,  we  have 

am=  a  •  a  •  a  .  .  .  to  m  factors  ; 

and        an=  a  •  a  '  a  .  .  .ton  factors. 

.  • .  am  x  an = («  •  a  •  a  .  .  .to  m  factors)  x(a-ci'a .  .ton  factors) 

=  a  •  a  ■  a  .  .  .  to m-\-n factors 

=  am  +  n  by  definition  (Art.  12) (1) 

. ,        _         „       am       a  •  a  •  a  .  .  .  to  m  factors 

Also  am  -i-  an  =  —  =  

an       a  •  a  •  a  ...  to  n  factors 

=  a -a-  a  .  .  .  to  m  —  n  factors 

=  a"'-'1 (2) 

From  Art.  103  we  have 

(a"T  =  «™ (3) 

and  am  X  bm  =  {ab)m (4) 

These  four  fundamental  laws  of  combining  exponents  are 
proved  directly  from  a  definition  which  has  meaning  only 
when  the  exponents  are  positive  and  integral. 

114.  Fractional  Exponents.  —  It  is  often  found  con- 
venient to  use  fractional  and  negative  exponents,  such  as  a*, 
a-5,  which  at  present  have  no  intelligible  meaning,  because 
we  cannot  write  a  \\  times  or  —5  times  as  a  factor.  It  is 
very  important  that  Algebraic  symbols  should  always  obey 


224  FRACTIONAL   EXPONENTS- 

the  same  laws  ;  and  to  secure  this  result  in  the  case  of 
exponents,  the  definition  should  be  extended  so  as  to  include 
fractional  and  negative  values.  Now  it  is  found  convenient 
to  give  such  definitions  to  fractional  and  negative  exponents 
as  will  make  the  relation 

am  x  an  =  am  +  n (1) 

always  true,  whatever  m  and  n  may  be. 

To  find  the  meaning  of  a*. 

Since  (1)  is  to  be  true  for  all  values  of  ra  and  n,  we  must 
have 

ca  x  cP  =  ct2  +  2  =  a1  ==  a. 

Thus  oh  must  be  such  a  number  that  its  square  is  a.  But 
the  square  root  of  a  is  such  a  number  (Art.  13).     Therefore 

a*  =  >/a (2) 

To  find  the  meaning  of  a*. 

By  (1)      a$  x  ah  X  oh  =  as  +  i  +  l  =  a1  =  a. 

Hence  as  must  be  such  a  number  that  when  taken  three 
times  as  a  factor  it  produces  a  ;  that  is,  as  must  be  equivalent 
to  the  cube  root  of  a. 

.-.     a*  ==  Va (3) 

To  find  the  meaning  of  a*. 

By  (1)  a*  x  at  x  at  x  a*  =  a3. 

...     at  =  Va5     .     .     .     .   (4) 

To  find  the  meaning  of  an,  where  n  is  any  positive  integer. 

%  (1) 
a  »  x  a*  x  a"  X  .  .  .  to  n  factors  =  a  «  +  » + »+-  •  *" tcrU19  =  a1  =  a  ; 

therefore  a"  must  be  such  that  its  nth  power  is  a. 

.-.     a"  =  Va (5) 

To  /t/icZ  Me  meaning  of  a",  w/tere  m  a?ia*  n  are  any  positive 
integers. 


NEGATIVE   EXPONENTS.  225 

By  (1)      a"  x  a"  x  a"  x to  n  factors 

—  +  — I 1- ton  terms 

=  a"    "    "  =  am 

m 

therefore  a"  must  be  equal  to  the  nth  root  of  am  ;  that  is, 

m 

an  =  vV* (6) 

ill  5 

Also,  «n  x  a"  x  a*  X  ......  to  m  factors  =  an  ; 

therefore  a"  rueaus  also  the  mth  power  of  a" ;  that  is,  from 
(5) 

m 

a'  =  (VST (7) 

m 

Therefore  from  (6)  and  (7),  a"  =  V«m  =  (Va)m  ...   (8) 

m 

Hence  an  means  the  nth  root  of  the  mth  power  of  a,  or  the 
mth  poiver  of  the  nth  root  of  a;  that  is,  in  a  fractional 
exponent  the  numerator  denotes  a  power  and  the  denominator 
a  root. 

Examples.  o:r  =  \Jxh,  a'§  =  \la5,  4§  =  V?  =  V^i  =  8. 

115.  Negative  Exponents.  —  (1)  To  find  the  meaning 
of  a\ 

By  (1)  of  Art.  114,  a0  x  a"  =  a0  +  n  =  a"; 

.-.     a0  =  an  -7-  an  =  1     .     .     .   (1) 

Hence,  any  number  whose  exponent  is  zero  is  equal  to  1. 
(See  Art,  45). 

(2)  To  find  the  meaning  of  a_n,  ivhere  n  is  any  positive 
number. 

By  (1)  of  Art.  114,  an  X  a~n  =  an~n  =  a0  =  1  [from  (1)]. 

Hence  an  =  ,     and     a~n  =  —  .     .     .     .    (2) 

a—  an  v 

Tints  ive  see  that  any  quantity  may  be  changed  from  the 

numerator  to  the  denominator,  or  from  the  denominator  to 

the  numerator,  of  a  fraction,  if  the  sign  of  its  exponent  be 

changed. 


226  EXAMPLES. 


Examples.  x~2  =  —; =  x*  =  \/x;  — -  =  x%  =  Vza; 

—  -  aWx-iy-3  =  - 


xy3  a~2b~sxy3 

27-i  =  -L  =  J=  =  -L  =  I  =  |  (by  (8)  of  Art.  114). 
2?l       V272       V36       32 

Otherwise  thus:  -L  =  J=-a  =  1  =  f 

(3)   To  _p?'o-ye  i/tai  am  -j-  an  =  am_n  /or  aM  va^es  o/  m 
cmd"  n. 

am  -7-  a*  =  —  =  am  x  a~n  =  am_n,  by  the  fundamental 

law. 

Examples,  a8  -r-  a6  =  a3-8  =  a-2  =    -. 

a2 

a  -7-  a~*  ==  a1  +  5  =  a5  . 

2«*  X  al  X  6a-$  =  j^+l-J+f-l  =  |a-i  =  ±. 
9a-^  X  a*  3a 


'a;3  x 


V?/~~2  x  Vx^       y~*  Xx* 


x*  x  y\  =  rf-ljl+t  =  afy 


?/• 


Note.  — It  appears  that  it  is  not  absolutely  necessary  to  introduce 
fractional  and  negative  exponents  into  Algebra,  since  they  merely 
supply  us  with  a  new  notation  in  addition  to  one  we  already  had.  It 
is  simply  a  convenient  notation,  which  the  student  will  learn  to 
appreciate  as  he  proceeds. 

EXAMPLES. 

Express  with  positive  exponents : 

3.    a^xaClxVS.^m.8lyl. 

^"8 


1.    2x    *a    3.  Ans.  -7  , 

x*a* 

2     2a>*  x  3a:-1  G 


V*8 


4.    Va-«  -h  Va7. 


TO  PROVE  THAT  (am)n  =  amn  JS  UNIVERSALLY  TRUE.    227 
Express  with  radical  signs  : 


5.  halx    *b    *.        Ans.  -=. 

6.  o~*  X  2a"i  -L 


7.    a;    3  _^  2a    *.     Ans. 


8.    7q~*  X  3a" 


4K* 


116.  To  Prove  that  (am)"  =  amn  is  Universally  True 
for  All  Values  of  m  and  n. 

1.    Ze£  n  6e  a  positive  integer,  and  m  have  any  value. 
Then  from  the  definition  of  a  positive  integral  exponent 
(am)n  =  am  x  am  x  am  x to  n  factors 


m  +m  +  m+ ton  terms 


=  a' 

=  amn (1) 

2.  Let  n  be  a  positive  fraction  *-,  ivhere  p  and  q  are 

positive  integers,  and  m  unrestricted  as  before.     Then 

p          

(am)n  =  (am)q  =  V(am)p  (Art.  114) 

=  9va^  by  (1) 

mp 

=  aT  (Art.  114) 

=  amn (2) 

3.  Let  n  be  negative,  and  equal  to  —p,  where  p  is  a  positive 
integer,  and  m  unrestricted  as  before.     Then 

1 


(a*)"  =  (am)~p  = 


(ory 


(Art.  115) 


-^  by  (1)  and  (2) 


=  a_rnp  ==  an 


(3) 


Hence  (am)n  =  (an)m  =  amn  for  cdl  values  of  m  and  n. 


4.    Let  n  = 


Then  we  have 


(am)~»  =  (a")m  =  a* 


(4) 


228     TO  PROVE  THAT  (ab)n  =  anbn  FOR  ANY  VALUE  OF  71. 

That  is,  the  nth  root  of  the  mth  power  of  a  is  equal  to  the  mth 
power  of  the  nth  root  of  a. 

5.    Let  m  =  —  and  n  =  -. 
m  n 

(am)»  =  (a")"  =  a™ (5) 

That  is,  the  nth  root  of  the  mth  root,  or  the  mth  root  of  the  nth 
root  of  a  is  equal  to  the  mnth  root  of  a. 

Examples.   (&!)*  =  b*x%  =  &*. 

(ifsy  =  (3*)8  =  3*  =  y/3. 

tySW  =  [(27a3)*]*  =  [(27«3)3]^  =  \ftoe. 

117.  To  Prove  that  (ab)n  =  anbn  for  Any  Value  of 

n.  —  This  has  already  been  shown  to  be  true  when  n  is  a 
positive  integer  (Art.  103). 

1.  Let  n  be  a  positive  fraction  ±_,  where  p  and  q  are 

9 
positive  integers.     Then 

V 

(ab)n  =  (ab)q. 
p 
Now  \_{ab)qY  =  (ab)p  (Art.  116) 

=  a*bp  (Art.  103) 

p  p 
=  (aqbq)q. 

...     (ab)q  =  aW (1) 

2.  Let  n  have  any  negative  value,  say  — r,  where  r  is  a 
positive  integer.     Then 

(ab)n  =  (ab)~r  =  —1— 

arb' 
which  proves  the  proposition  generally. 


EXAMPLES. 


229 


In  this  proof  the  quantities  a  and  b  are  wholly  unrestricted, 
and  may  themselves  have  exponents. 

P  _  ! 
Let  —  —  -.     Then  from  ( 1 )  we  have 
q       n  v   ' 


(ab)n  =  anbn. 
.'.     \ab  =  Va  •  y& 


(3) 


That  is,  the  nth  root  of  the  product  is  equal  to  the  product 
of  the  nth  roots. 


EXAMPLES. 

1.    (e%~2)3  ^_  (#22/-1)-^  ==  x*y~$  -7-  x~%jp 


x$y~\ 


Express  with  positive  exponents 


/27^\-§ 
\8a-*) 

farjV2 

W)    ' 


Ans. 


2x*y* 

4 
9aV" 

16ac4. 


5.    (xay-bY(xY)~a- 


Ans. 


6.  VxVx-K 

7.  (4a~2  --  9«2)-i 


8.    Vab-^-'2  x  (a-^-2c-4)-i 


9.    v'a46^6  X  (cAb-1) 


10.    (a-^)~s  X  V^Va^6. 


11.    V(a  +  &)5  X  (a  +  &)~~§. 


a  +36 

3cwc 
2  " 


a?" 
a  +  6. 


Rem.  —  Since  the  laws  of  the  exponents  *  just  proved  are  universally- 
true,  all  the  ordinary  operations  of  multiplication,  division,  involution, 
and  evolution  are  applicable  to  any  expressions  which  contain  frac- 
tional and  negative  exponents. 


Called  the  index  laws. 


230  EXAMPLES. 

The  reason  for  the  arrangement  in  Ex.  3,  Art.  107,  may 
now  be  seen.     Thus  the  descending  powers  of  x  are 

ryO         /yi2  /y»  "I        ^_  

.*/     ,      Jj     ,      ^/,        J.   ,         ,  ,  , 

X     XJ     X* 

as  may  be  seen  (Art.  115)  by  writing  the  terms  as  follows : 


/y»3  /yi2  /-v^l  /y»0  ^yi  1  /y>  — 2  rv%  -**    3 


12.    Multiply  3a;- 1  +  a;  +  2xf  by  x?  —  2. 

Arrange  in  descending  powers  of  x. 
x    +  2xf  -J-  3x~^ 


a*  -  2 

jcf  4.  2x    +  3 
—  2x    —  4^3  - 

-  6x"4 

xf  —  4x^  +  3      —  6x"i. 

13.  Divide  3x*y~3-\-x%  —  Sxsy~i  — y~\  by  x^+y~*  —  2x%~"s. 
Arrange  in  descending  powers  of  x. 

X?  —  2X6 y~\  -{-  y~\  |  aji  —  3X3?/-^  +  3X6?/ -4  _  y~\  I  x^  —  i/-^ 
X?  —  2x%~^  +     Xe?/-^" 

—  xsy~6  4-  2x6?/  -3  —  ?/-i 

—  x%~~£  4-  2x%~s  —  ?/-i 

Multiply 

14.  x*  -f  ?/t  by  x^  —  ?/^.  Ans.  xt  —  ?/i. 

15.  x4  -h  x2  4-  1  by  x"4  —  x"2  4-1.  x4  +  1  4-  #~4. 

16.  a-i  4-  «"^  +  1  by  a~3  -  1.  a"1  -  1. 
Divide 

17.  21x  4-  a#  4-  srf  4-  1  by  3x?  +  1.  7x§  —  2x£  4-  1. 

18.  15a-3a^-2a-H8a-1by  5af+4.  3a^-Scr^  +  2a'1. 
Find  the  square  root  of 

19.  9x  —  12x2  4-  10  —  4x"2  4-  x"1.  3x?  —  2  4-  %~^ 

20.  4x"  4-  Vx-™  +  28  —  24x-*  -  lGx*.  2x"z  -  4  4-  3x~z. 


SURDS  —  DEFINITIONS.  231 

SURDS    (RADICALS). 

118.  Surds.  Definitions.  —  When  the  indicated  root  of 
a  quantity  cannot  be  exactly  obtained,  it  is  called  an  irra- 
tional quantity  or  a  Surd. 

Thus,  v'2,  ?4,  Va®,  V«2  +  b\  at,  are  surds. 

When  the  indicated  root  can  be  exactly  obtained,  it  is 
called  a  rational  quantity.  Thus  y.i*6,  V9,  v«4?  are  rational 
quantities,  though  in  the  form  of  surds. 

The  order  *  of  a  surd  is  indicated  by  the  index  of  the  root. 
Thus  y$  fa  are  respectively  surds  of  the  third  and  fifth 
orders. 

The  surds  of  the  most  common  occurrence  are  those  of 
the  second  order ;  they  are  sometimes  called  quadratic  surds. 
Thus  03,  *Ja,  \lx  +  y  are  quadratic  surds.  Surds  of  the 
third  and  fourth  orders  are  called  cubic  and  biquadratic  surds 
respectively. 

When  the  same  root  is  required  to  be  taken,  the  surds  are 
said  to  be  of  the  same  order.  Thus,  ya,  Va  -f-  6,  and  5s  are 
all  surds  of  the  third  order  or  cubic  surds. 

Surds  are  said  to  be  like  or  similar  when  they  are  of  the 
same  order,  or  can  be  reduced  to  the  same  order,  with  the  same 
quantity  under  the  radical  sign. 

Thus,  4<Sl  and  os/l  are  like  or  similar  surds ;  also  5^2  and 
SVTG  are  like  surds  ;  2^3  and  3^2  are  unlike  surds. 

A  mixed  surd  is  the  product  of  a  rational  factor  and  a 
surd  factor.     Thus  4^5,  and  3^7  are  mixed  surds. 

When  there  is  no  rational  factor  outside  of  the  radical 
sign,  the  surd  is  said  to  be  entire.  Thus  V^2  and  V^3  are 
entire  surds. 

The  rules  for  operating  with  surds  follow  from  the  propo- 
sitions of  the  preceding  Articles  of  this  Chapter. 

*  Sometimes  called  degree. 


232        TO  REDUCE  AN  ENTIRE   TO  A  MIXED   SURD. 

119.  To  Reduce  a  Rational  Quantity  to  a  Surd 
Form.  —  It  is  often  desirable  to  write  a  rational  quantity  in 
the  form  of  a  surd.     Thus 

a  =)Ja2  =  y/^  =  V^5    3  =  fi  -  V^7- 
In  the  same  way  the  form  of  any  surd  may  be  altered. 
Hence  the 

Rule. 

Any  rational  quantity  may  be  expressed  in  the  form  of  a 
surd  of  any  required  order  by  raising  it  to  the  poiver  corre- 
sponding to  the  root  indicated  by  the  surd,  and  prefixing  the 
radical  sign. 

Examples.  5  =  0!5  -  \/i25  =  y/?. 

(a  +  b)  =  y/(a  +  by  =  V(«  +  &)3- 

120.  To  Introduce  the  Coefficient  of  a  Surd  under 
the  Radical  Sign.  —  We  have 

3^2  =  y/9  x  \/2  (Art.  119) 

=  \/9  x  2  [(3)  of  Art.  117]  =  ^18. 

2y/5  =  V25  X  V*  =  V23  X  5         =  V^O- 
x\2a  —  x  =  \2ax2  —  xz. 
Rule.  Reduce  the  coefficient  to  the  form  of  the  surd  and 
then  multiply  the  surds  together, 

EXAMPLES. 

Express  as  entire  surds. 


1.  lly/2.        Ans.  y/242. 

2.  5y/6.  YnO. 


3.  14y/5.  Ans.  \/980. 

4.  6'V5.  V^£ 


121.  To  Reduce  an  Entire  to  a  Mixed  Surd. — 

We  have 

y/32  =  y/l6  x  2  =  y/To  x  y/2  =  4y/2  ; 

also  ]/M  =  y/"6  X  V&  =  a,V*« 


REDUCTION   OF   SURDS    TO   EQUIVALENT   SURDS.     233 

Rule.  Resolve  the  quantity  under  the  radical  sign  into  two 
factors,  one  of  which  is  the  greatest  perfect  power  correspond- 
ing to  the  root  indicated;  extract  the  required  root  of  this 
factor,  and  prefix  the  result  as  a  coefficient  to  the  indicated 
root  of  the  other. 

When  a  surd  is  so  reduced  that  the  smcdlest  possible  integer 
is  left  under  the  radical  sign,  it  is  said  to  be  in  its  simplest 
form.     Thus, 


The  simplest  form  of  \/l28 

=  034  x  2  = 

8V2. 

EXAMPLES. 

Express  in  the  simplest  form  : 

1.    \'288.        Ans.  12y/2. 

3.    v/SGol 

Ans.  6a\a. 

2.    V1029-                  7V3- 

4.    \j'21a*b\ 

3ab2\j3a~b. 

122.  Reduction  of  Surds  to  Equivalent  Surds. — 

To  reduce  surds  of  different  orders  to  equivalent  surds  of  the 
same  order. 

For  example,  take  y/5  and  \/ll. 

sji  =  oi  =  bi  =  yr*  =  yWo. 

yn  =  ii*  =  nf  =  yir2  =  yiii. 

In  general,  let  n^am  and  g\bp  be  two  surds  of  different 
orders.  Then  we  have  to  change  these  into  equivalent  surds 
whose  fractional  exponents  have  the  same  denominator. 

We  can  reduce  both  surds  to  the  order  nq  as  follows : 


and  SjbP  =  ¥  =  &*"  =  n  \/V\ 

Thus  the  equivalent  surds  of  the  same  order  are  "\/am9  and 

Rule.  Represent  the  surds  with  fractional  exjwnents;  re- 
duce these  fractions  to  their  least  common  denominator;  then 
express  the  resulting  fractional  exponents  with  radical  signs, 


234 


ADDITION  AND   SUBTRACTION   OF  SURDS. 


and  reduce  the  expressions  under  the  radical  signs  to  their 
simplest  forms. 

Note.  —  In  this  way,  surds  of  different  orders  may  be_coinpared. 
Thus,  if  we  wish  to  know  which  is  the  greater,  >fb  or  f/ 11,  we  have 
only  to  reduce  them  to  the  same  order,  as  above;  we  see  that  the 
former  is  greater  because  125  is  greater  than  121. 

EXAMPLES. 

Express  as  surds  of  the  twelfth  order,  with  positive  ex- 
ponents : 


1.  xk  Ans.  1y^  4.    cta. 

1 

2.  a-1  -r-  a  *.  ii7=- 

ya6 

3.  -3-  *V*. 
a~*  v 

Express  as  surds  of  the  same  lowest  order  : 


Ans.    y/x9. 


x"y. 


5.   yafyi. 

6.  yss  x  v«-1^"2-  v  ^ 


7.  \/a,]/a*.  Ans.'y/a^^a10. 


10.  y^6*,Va&.    t^,*^. 


1 1 .  Which  is  the  greater  y/l4  or  ^6  ?  ^6. 

123.  Addition  and  Subtraction  of  Surds.  — Let  it 

be  required  to  find  the  sum  of  \Jl2,  ^75,  —  \A±8,  and  ^50. 
Here  we  have  (Art.  121) 

V/l2  +  y/75  -  y/48  +  ^50  =  2\/3  +  5^3  -  4^3  +  5^2 

=  (2  +  5  -  4)y/3  +  5^2 

=  305  +  5y/2. 
Rule.  Reduce  the  surds  to  their  simplest  form;  then  add 
or  subtract  the  coefficients  of  similar  swds  and  jwejix  the  result 
to  the  common  surd,  and  indicate  the  addition  or  subtraction 
of  unlike  surds. 
Thus,  3^20  +  4y/5  +  y/j  +  V^5 

=  Gy/5  +  4\/5  +  Jy/5  +  5^5 
=  lO-Jy/5  +  50$. 


MULTIPLICATION   OF  SURDS.  235 

EXAMPLES. 

Find  the  value  of  the  following : 

1.  3v/45  +  7v/5  -  v/20.  Ans.  U\fi. 

2.  4^63  ■+■  5^7  -  80*8.  ft. 

3.  y/44  -  5y/l76  +  2^99.  -12\/n. 

4.  2y/363  -  5\/243  +  ^192.  -  15^3. 

5.  2VJ  +  8VS-  3V2. 

6.  V40  -  iV^20  +  V135-  3V5- 
124.  Multiplication  of  Surds. —  (1)  When  the  surds 

are  of  the  same  order. 

To  multiply  a  n^x  by  b  \y. 

_  11 

Here     a\x  x  bn\/y  =  cra^  x  %"  [Art.  114,  (5)] 

1  1  1 

=  afruy1  =  ab{xy)n  (Art.  117) 

=  abn^xy. 
(2)  TPVien  £/*e  surc7s  are  o/  different  orders. 

To  multiply  a  \x  by  6  y^. 

_  _  i  i 

Here  ayx  x  b\jy  =  axn  x  fo/m 

m       n 

=  abxmny™  (Art.  122) 

=  ab(xmyn)™  (Art.  117) 

=  aft  m\Jxmyn. 

Rule.  TFAen  Me  surds  are  of  the  same  order,  multiply  sep- 
arately the  rational  factors  and  the  irrational  factors.  When 
the  surds  are  of  different  orders,  reduce  them  to  equivalent 
surds  of  the  same  order,  and  proceed  as  before. 

Thus,  3^2  x  7y/6  =  21^12  =  42^3. 

5^2  x  2y/5  =  5V*  X  2^ 
=  106v/500. 
A  compound  surd  is  an  expression  involving  two  or  more 


236  MULTIPLICATION    OF  SURDS  —  EXAMPLES. 

simple  surds.  Thus  20i  —  30>,  and  y/a  -f-  V&  are  compound 
surds. 

The  multiplication  of  compound  surds  is  performed  like 
the  multiplication  of  compound  Algebraic  expressions. 

Multiply  20c  -  5  by  30c. 

The  product  =  30c(20c  —  5)  =  Gx  —  150c. 

Note.  —  To  multiply  a  surd  of  the  second  order  by  itself  is  simply 
to  remove  the  radical  sign ;  therefore  \Jx  x  sjx  =  x. 

Multiply  605  -  50>  by  20$  -f  302. 

The  product  =  (60$  -  50!)(20$  +  302) 

=  36  +  18  ^6  -  10\/6  -  30  =  6  +  803. 
The  following  case   of    the    multiplication    of   compound 
surds  deserves  careful  attention.     The  product  of  the  sum 
and  difference  of  any  two  quadratic  surds  is  a  rational  quan- 
tity.    Thus 

(303  +  40$)  (303  -  403)  =  (305)2  -  (403)2 
==  45  -  48  =  -3. 
Also  (0z  +  00  (0^  -  V?)  =  (00s  -  (0O2  =  a  -  b. 
A  binomial  in  which  one  or  both  of  the  terms  are  irra- 
tional, is  called  a  binomial  surd. 

When  two  binomial  quadratic  surds  differ  only  in  the  sign 
which  connects  their  terms,  they  are  said  to  be  conjugate. 
Thus  _ 

303  +  4^3  is  conjugate  to  3^5  —  4^3. 


Similarly,  a  —  0t'2  —  b2  is  conjugate  to  a  +  0r  —  b 
The  product  of  two  conjugate  surds  is  always  rational 

EXAMPLES. 

Find  the  value  of 


1.  20U  x  0*1.     Ans.  1405. 

2.  304  x  05.  120$. 

3.  20L5  x  30L  3003. 


4.  y/l^xVl47.  Ans.  1480). 

5.  03  x  \fi.  V^8- 

6.  VixVlxVixVif     V* 


TO  RATIONALIZE  THE  DENOMINATOR  OF  A  FRACTION.    237 

7.  (3y/ic  —  5)  x  2y/#.  ^dns.  6x  —  10y^.  . 

8.  (\?x  —  y/a)  x  2^.  2a;  —  2^ax. 

9.  (V^7  +  5v/3)(2v/7  -  403).  6^21  -  46. 

10.  (3\/5  -  4y/2)(2v/5  +  3V2).  6  +  y/lO. 

11.  (5  +  3y/2)  (5  -  3\/2).  7. 

12.  (3^  +  ^  -  9a)  C3^  -  V^  -  9a).  18a  -  x. 

125.  To  Rationalize  the  Denominator  of  a  Frac- 
tion. —  The  process  by  which  surds  are  removed  from  the 
denominator  of  any  fraction  is  known  as  rationalizing  the 
denominator. 

(1)  Wlien  the  denominator  is  a  monomial. 

_2_  _        2y/3         =  2^/3 

y/3      y/3  x^a        3 

i3/?  =  4s/2  x  9  =  ^/H  =  v51 

V  3       V  3  x  9       V  27         3  ' 

Rule.  Multiply  both  terms  of  the  fraction  by  any  factor 
which  will  render  the  denominator  rational. 

(2)  When  the  denominator  is  a  binomial  quadratic  surd. 

b2 


Rationalize  the  denominator  of 


\'a2  +  b2  +  a 


The  expression  =  h* x  ^  +  *  ~  a 

\Jar  +  b2  +  a       ija2  +  b2  —  a 

(a2  +  6-)  —  a2 
Rule.    Multiply  both   numerator  and   denominator  of  the 
fraction  by  the  surd  which  is  conjugate  to  the  denominator. 

When  the  denominator  of  the  fraction  is  rationalized,  its 
numerical  value  can  be  more  easily  found.    Thus,  the  numeri- 

/-  2 

cal  value  of  |y3  can  be  found  more  easily  than  that  of  -y=- 


238 


DIVISION   OF  SURDS. 


29 

Given  y/o  =  2.2360G8,  find  the  value  of  = ^7=- 

■  ~~  ^y/5 

It  might  seem  at  first  sight  that  we  must  subtract  twice 
the  square  root  of  5  from  7,  and  divide  29  by  the  remainder 
—  a  troublesome  process,  as  the  divisor  would  have  7  figures. 
We  may  avoid  much  of  this  labor  by  rationalizing  the  de- 
nominator.    Thus, 

29        =  29(7  +  2y/5) 
49  -  20 


2y/5 


7  +  2y/5 


=  11.472136. 


EXAMPLES. 

Rationalize  the  denominators  of 


1.4. 

2.  vi. 


VI- 

2  +  V^5 
^5  -  l" 


Aiis.  jV15- 

iV/6. 

iVw- 

3^5 


7  + 


0. 


7^3-5^2 


7.   10Vg  ~  2Vg.      8  -  v^i. 


3^6  +  2^7 

y/7  +  y/2  _ 
9  +  2y/u' 


V?-^ 


126.  Division  of  Surds.  —  Since  a n^x  X  bn^y  —  ab \xy 
(Art.  124),  therefore 

abnyjxy  -~-  ayfi  =  bn}Jy. 
Rule.  When  the  surds  are  of  the  same  order,  divide 
separately  the  rational  factors  and  the  irrational  factors. 
When  the  surds  are  of  different  orders,  reduce  them  to 
equivalent  surds  of  the  same  order,  and  proceed  as  before. 
Then  the  denominator  may  be  rationalized  (Art.  125). 

Thus,  4^75  -*-  25^56  =  ^L  =    4  X  5V^L 

25030        25  x  2^14 


21/3      2*  I  ]± 

—  -&VT4  —  syyy 


X   14 
X  14 


012 
35' 


BINOMIAL    SURDS  —  IMPORTANT  PROPOSITIONS.     239 

The  only  case  of  division  of  a  compound  surd  which  we 
shall  consider  is  that  in  which  the  divisor  is  a  binomial 
quadratic  surd.  We  may  express  the  division  by  means  of 
a  fraction,  and  then  practically  effect  the  division  by  ration- 
alizing the  denominator.     Thus, 

Divide  ^3  +  sjl  by  2^3  -  ^2. 

The  quotient  =    V^_+  ^  =    (V^  +  V^)  (2^3  +  y/2) 
2y/3  -  \J2        (2\/3  -  y/2)(2\/3  +  ^2) 

=  8  +  3y/6  =  8  +  3y/6 
12-2  10      * 


EXAMPLES. 

Find  the  value  of 

1.    21^384  --  8v/98. 

Ans.  3y/3. 


2.    5^27-^-3^24. 


3.  -13^125  --  oy/65.  _ 
Ans.  —^13. 

4.  6y/l4  -h  2^21.  ft. 


11  -  3y/7 

2 
19  -  Cy/2 

17 

2  +  Vfa 


5.  29  -=-  (11  +  3^7). 

6.  (3y/2  -  1)  -  (3y/2  +  1). 

7.  (2^3  +  7^2)  -=-  (5y/3  -  4y/2). 

8.  (2a  -  fa)  -  (2\/^-  2/). 

127.  Binomial  Surds.    Important  Propositions. 

(1)    The  square  root  of  a  rational  quantity  cannot  be  partly 
rational  and  partly  a  quadratic  surd. 

If  possible,  let  ya  =  b  +  \/c. 

Squaring,  we  have    a  =  B?  +  26y/c  -f  c. 

V  26 

that  is,   a  surd   is   equal   to  a  rational  quantity,    which  is 
impossible. 


240  SQUARE  ROOT   OF  A   BINOMIAL   SURD. 

(2)  In  any  equation  consisting  of  rational  quantities  and 
quadratic  surds,  the  rational  parts  on  each  side  are  equal, 
and  also  the  irrational  parts. 

If  x  -f-  yjy  =  a  +  \/6,  then  will  x  =  a,  and  y  =  b. 

For  if  x  is  not  equal  to  a,  suppose  x  =  a  +  m ; 
then  a  +  m  +  y/y  =  a  -f  y/&; 

that  is,  m  +  \Jy  =  \/b, 

which  is  impossible  by  (1).     Hence  x  =  a,  and  therefore 
y/y  =  \lb. 

Note.  — When  z  +  \[y  =  a  +  \^>>  we  can  conclude  that  x  =  a  and 
>Jy  z=  \[b  only  when  VV  and  0>  are  really  irrational.  We  cannot,  for 
example,  from  the  relation  6  +  ^4  =  5  +  y/9,  conclude  that  6  =  5  and 


(3)  If  Va  +  Sib  =  sfx  +  )/y,  then  Va  -  \lb  =  sjx  -  \[y. 
For  by  squaring  the  first  equation  we  have 
a  +  sib  =  a;  +  y  +  2^xy. 
.-.     a  =  x  -\-  y,  and  y/&  =  2^xy. 
Subtracting,  a  —  ^b  =  x  —  2\jxy  +  ?/ ; 

.-.     \la  -  ^b  =  tfx  -  Vy. 

128.  Square  Root  of  a  Binomial  Surd. —  The  square 
root  of  a  binomial  surd,  one  of  ivhose  terms  is  rational,  may 
sometimes  be  expressed  by  a  binomial,  one  or  each  of  ivhose 
terms  is  a  quadratic  surd. 

Let  a  -f-  \/&  be  the  given  binomial  surd. 

Assume  V^a  +  V&  =  >fx  +  V^-      •     •     •  0) 

By  (3)  of  Art.  131,     \la  -  \/&  =  V^  -  V&      ...  (2) 

Multiplying  (1)  by  (2),v/cr  -  6  =  x  -  y (3) 

Squaring  (1)  a  +  \lb  =  x  +  2\l  xy  +  y.      .  (4) 

Therefore  by  (2)  of  Art.  127,  a  =  x  +  y (5) 


SQUARE  ROOT  OF  A    BINOMIAL   SURD.  241 

Hence,  from  (3)  and   (5),  by  addition  and   subtraction,  we 
have 


"  +  y  ~  \ (6) 


y  -  °  ~  y  ~  \ (?) 

which  substituted  for  x  and  y  in  (1)  and  (2)  will  give  the 
values  of  \  a  4-  \Jb  and  SI  a  —  ^b. 

Find  the  square  root  of  16  -f-  2y/o5. 

Here  a  ==  16,  and  ^6  =  20*5. 
Then  aa  -  b  =  256  -  220  =  36, 

which  in  (6)  and  (7)  gives 

x  =  |(16  +  6)  =  11. 
y  =  £(16  -  6)  =     5. 
Hence  Y^16  +  2^55  =  V^  -f  Vo. 

From  the  values  of  x  and  y  in  (6)  and  (7),  it  is  clear  that 
each  of  them  is  itself  a  complex  surd  unless  \a2  —  b  is 
rational ;  and  the  expression  \Jx  +  \y  will  be  more  compli- 
cated than  V  a  +  \Jb  itself.  Hence  the  above  method  for 
finding  the  square  root  of  a  +  y/&  fails  entirely  unless  a'2  —  b 
is  a  square  number ;  and  as  this  condition  is  not  often  satis- 
fied, the  process  has  no  great  practical  utility. 

TJie  square  root  of  a  binomial  surd  may  often  be  found  by 
inspection.  For  we  see  from  (4)  and  (5)  that  we  have  to 
find  two  numbers  whose  sum  is  a  and  whose  product  is  b ; 
and  if  two  rational  numbers  satisfy  these  conditions,  they 
can  generally  be  found  at  once  by  inspection.     Thus 

1.    Find  the  square  root  of  11  +  2^30. 

We  have  only  to  find  two  numbers  whose  sum  is  11,  and 
whose  product  is  30j  and  these  are  evidently  6  and  o. 

Hence  11  +  2^30  =  6  +  2\/6  x  5  +  5 

_  =  (y/6  +  foy.  __ 

.*•     y^  +  V5  =  the  square  root  of  11  +  2^30. 


242  EQUATIONS  INVOLVING   SURDS. 

2.    Find  the  square  root  of  53  —  12^10. 

We  must  write  the  binomial  so  that  the  coefficient  of  the 
surd  is  2.     Thus 

53  -  12y/l0  =  53  -  20560. 

The  two  numbers  whose  sum  is  53  and  whose  product  is 
360  are  45  and  8. 

Hence       53  -  2y/360  =  45  -  2^45  x  8  +  8 

.-.     ^53  -  12V/10  =  \/45  -  V^8  =  3V^5  -  2^2. 

EXAMPLES. 

Find  the  square  root  ot 

1.  7  +  2\/l0.  Ans.  \/5  +  \fi. 

2.  13  +  2y/30.  VlO  +  y/3. 

3.  5  +  2y/6.  \/J  +  V^« 

4.  47  -  4y/33.  2^11  -  *fo 

5.  15  +  2y/56.  V^8  +  y/7. 

The  cube  root  of  a  binomial  surd  may  sometimes  be  found 
by  a  method  similar  to  the  one  just  given  for  obtaining  the 
square  root.  But  the  method  is  very  imperfect,  and  is  of  no 
practical  importance. 

129.  Equations  Involving  Surds.  —  Equations  some- 
times occur  in  which  the  unknown  quantity  appears  under 
the  radical  sign.  In  the  solution  of  such  equations,  special 
artifices  are  often  required.  We  shall  here  consider  only  a 
few  of  the  easier  cases,  which  reduce  to  simple  equations. 
These  can  generally  be  solved  by  the  following 

Rule.  Transpose  to  one  member  of  the  equation  a  single 
radical  term  so  it  will  stand  by  itself;  then  on  raising  each 
member  to  a  poiver  of  the  same  degree  as  the  radical,  it  will 
disappear.  If  there  are  still  radical  terms  remaining,  repeat 
the  process  till  all  are  removed. 


EXAMPLES.  213 

EXAMPLES. 


1?  Solve  2\Jx  -  s/ix  -  11  =  1. 

Transposing,  2^x  —  1  =  ^ix  —  11. 

Squaring,  4x  —  4ysc  +  1  =  4a;  —  11. 

Transposing  and  dividing  by  —4,  yx  =  3.         .•.     x  =  9. 

/1y  Solve  SJx  —  \[\~^x  +  V7^  = 

Transposing,  \x  —  \1  —  x  = 


—  sfx. 

—  2^x  -f  a;. 

—  4y/a;  +  4a;. 
6a;. 


Squaring,                          x  —  \L--x  = 

Canceling  a;  and  squaring,  1  —  x  = 

Transposing  and  squaring,  25a^  = 

Dividing  by  25a;,  x  =  |§. 

When  radicals  appear  ji  a  fractional  form,  the  equation 
should  be  first  cleared  of  fractions  in  the  usual  way  before 
performing  the  involution. 

3.    Solve  6V^-U  =  2^  +  \ 
S^x  \x  +  6 

Clearing  of  fractions 

6x  +  2d\Jx  -  66  =  6a;  +  3\/x. 

.-.     22^  =  66.         .-.     x  =  9. 
Solve  the  following  equations. 
4.    y/a~7,  =  3#  ^HS.  14. 

^  \JAx  -  7  =  5.  33. 

(g)  y/5a>  -  1  =  2\Jx  +  3.  13. 

((7)  13  -  Voa?  -  4  =  7.  44. 

^  2\/3  -  7a;  -  3\/8a;  -  12  =  0,                                       {. 

d)  \1  +  V3  -f-  sjtx  =  2.  6. 


244 


EXAMPLES. 


EXAMPLES, 


1.    Multiply  a~*  by  a<L 

^4?is.  aX*. 

2.    Multiply  3a*&*  by  4a^. 

12afet 

3.    Divide  ai*  by  a;4. 

<A 

4.    Divide  a~2x~*  by  a-3. 

aa;-* 

Find  the  numerical  values  of 

5.    16" «.             u4ns.  -J. 

8. 

4-i. 

-4ns.  £. 

6.    27*.                        81. 

9. 

16~ 

'•                      J- 

7.    8i.                            4. 

10. 

(tts 

)"i.                25. 

Express  with  positive  exponents  : 

11.    sjazy  -+-  \/o?/*. 

Ans.  a%y^  +  a3?/ 

12.    yah?  +  \Ja^y. 

aw  -\-  a$y^ 

\Jv*        Va-1            1 

3 

5,1,   a?t 
a*        4 

/ '      '         3  /~          '          5  1 

yic-1          ya         4  y^,_ 

14.    (*  +  V»)"  +  V'j  x  V*   7 

15     /«~25\-3    .    /ab-1^5 

16. 


+ 


i  i 


17. 


18. 


/a-8\-|  ^  /'/a-^V^Y2 

V     At"2  *    B"  W 


4«  +  i 


(2")' 


(2"-') 


l\n  +  l 


2 


Express  with  radical  signs  and  positive  exponents : 

_  L 


19. 


2a  ~2       g~*        4.r^ 
a-f         3a         gj-i 


^l?is.  — -  H = 

y/a      3y/a 


+  7T=« 


20.   a*  x  a*  +  a~*  •*-  a   *". 


EXAMPLES. 


245 


21.    a^"1  -  a-5&. 


22.    a-sb~?  +  3a?6-t. 


,4ns. 


oV& 


1  -  xt. 

x2  -  l. 


Multiply 

23.  1  +  x$  +  x*  by  1  -  sb*. 

24.  x*  —  x$  4-  x^  —  af"£  by  a$  _|_  #1. 

25.  cc*  +  &*  4-  1  by  cc_n  +  a~£  +  1. 

^Ins.  af1  4-  2a£  +  3  +  2z-2  4-  »~*. 

26.  1  -  2s^x  -  2^  by  1  -  \Jx.       1  -  jc*  —  2»±  4-  2A 

27.  2  V«5  -  a*  -  -  by  2a  -  3y/i  -  a~i 

-4?is.  4a3  -  8a*  -  5  4-  10a~*  4-  3a~i 
Divide 

28.  16a~3  4-  6a"2  +  5a"1  -  6  by  2a"1  -  1. 

Ans.    8a~2  +  7a"1  4-  6. 

29.  21a3x  +  20  -  27ax  -  26a2x  by  3a*  -  5. 

Ans.  la2*  4-  3ax  —  4. 

30.  8c~n  -  8cn  4-  5c3*  -  3c~3"  by  5c*  -  3c-n. 

Ans.  c2"  —  1  4-  c"2*. 

31.  i  _  y/a  _  2a  4-  2a2  by  1  -  a*.  1  -  2a  -  2ak 

32.  x*  —  xy*  4-  oj¥^  —  y*  b}r  #?  —  y*.  x  4-  2/« 

33.  ai4-&l-c^4-2a£&*bya*4-&*4-ci      a*  4-  &*  -  c*. 
Find  the  values  of 


«•  [(4-*M^-") 


-4ns.  1  4-  (1  -  X2)*- 


35.  (a-^'-^a-iT)3.       aar34-  8a**-1  +  Ba'^x  4-  a"1**. 

36.  (la*  -  a"*)2.  JaS  -  |  4-  a~i 


37. 


38. 


a*  -  8ah 
«H  2V«6  4-  4^' 


a;  —  50c  —  14 


-  (■ + &'■ 


a^(ai  _  263). 
1. 


246 


EXAMPLES. 


2ix~la  +  16a-2a2, 


Find  the  square  root  of 

39.  4:X2a~2  -  Wxa-1  +  25  - 

Ans.  2xa~x 

40.  x?  —  4a5*»  +  Ax  -f-  2x*  —  4x*  +  a$.       Xs  —  2x*  -f  x 

41.  4a  -  12a^  +  96?  +  16aM  -  24&M  +  i6ci 

-4?is.  2a'  —  36'  +  4c^ 


3  +  4x~1a. 

4 


Express  as  entire  surds 
2a 


9a26 


43. 


2a  */27x\ 
3x\    a2  ' 


8ax. 


44 


45 


a;"V     2/3  V  y 


Express  in  the  simplest  form : 


46. 

3y/l50  and  2^720. 

J.ns.  15  V 

6  and  24^5 

47. 

V-108zy. 

—  3xy  \ix 

48. 

y/a3  +  2a26  +  ab2. 

(a  +  6)  y/a 

49. 

y8x4y  -  24xhf  +  24x2ys  - 

8x-?/4.           2 

(»  -  2/)V^ 

50. 

V768,  y/l250,  y/3888. 

*V5 

,  5y/2,  6y/3 

Express  as  surds  of  the  same  lowest  order : 

51. 

y/a#2,    y/a9x6. 

^bis.  2V«13^'265  Y"6^ 

52. 

y/5,  y/TT,  y/l3. 

Vi25, 

Vl21,y/l3 

53. 

Vs,  y/3,  Ve. 

v< 

54,  VSfi  V5 

54. 

3§,  2t,  5*. 

y6561,    yo 

12,  V15025 

Find  the  value  of 

55. 

y/243  +  y/27  +  ^48. 

^l?is.  16  y/3 

56. 

2  Vl89  +  3y/875  -  7y/56. 

7V7 

57. 

5y/81  -  7yi92  +  4y648. 

ii  v$ 

58. 

3*^162  -  7  y/32  +  V125u- 

0 

59.  3y/2  +  4y/8  -  y/32. 

60.  23y/4  +  5y/32  -  y/To8. 

61.  4y/l28  +  4^75  -  5y/l62. 


7y/2. 

9  V?- 
20^3  -  13y/2. 


EXAMPLES.  247 

62.  5y/T28  x  2y/432.  Ans.  2 AOy/ 4. 

63.  sijb  x  4y/a-  12  12y/«^"4. 

64.  ^2  X  V^  X  yi.  12y/648000. 

65.  4y/5  X  2 Vll-  8y/l5125. 

a;Y  a;          V  2a4  a 

67.  (2y/3  +  3v/2)2.  30  +  1205. 

68.  (y^  +  >Jx  -  l)\!x  -  1.  flj  -  1  +  \Jx2  -  x. 

69.  (y/z  -f  a  —  y/a;  —  a)^x  -fa.  a  -f  a  —  \Jx2  —  a2. 

70.  (y/2  +  V^3  -  \^)(V^  +  V/3  +  \/5).  2^6. 

71.  (Vl2  +  y/l9)(Vl2  -  y/19).  5. 

72.  (x2  +  x\/2  +  l)(x2  -  x\/2  +  1).  x4  +  1. 
Rationalize  the  denominator  of 

2j/£+3v/I  ^  3^2  _  2^, 
5  +  2^6 

y^5  + y/3 

75.    ?Jl^.  i(7  +  3y/5) 


73. 
74. 


76. 

77. 


f 


X2 


Six*  +  a2  +  a 

y/i  +  jg  -  yr 


X  — 

Sjx2  - 

-2/2 

\Jx2 

+  a2- 

—  a 

1  - 

-Vi- 

-    X* 

y/l   +  rf  +  y/l   _  ^  X2 

Find  the  numerical  value  of  the  following  to  five  places  of 
decimals : 

79.  ii,    i°  Ans.  9.8995,  3.77964. 

y/2    ^7 

80.  ^?,   _L.  .81649,  .28867. 

^3     2\/3 


248 


EXAMPLES. 


si.  vfc  +  0?. 

4  -f-  y/i5 
Find  the  value  of 

82.    *M  +  JL. 

2^98       7y/22 


Aras.   V5  -  -y/3  =  .50402. 


*§■ 


83      3018     .    6^84 
50L12    '    0592* 

*$■ 

84.    (3  +  VH)(v^5  -  2)  -s-  (5  -  03). 

5 

85       v/a      .  0*  +  0» 

y/a  —  y/a;             0b 

0x» 
a  —  a; 

Find  the  square  root  of 

86.  41  -  240*. 

87.  83  -f  12035. 

88.  101  -  28013. 

89.  117  -  36010. 

90.  280  +  560*1. 

91.  8  +  405. 

92.  4  -  v/l5. 

93.  75  -  12021. 

^Iws.  40*  -  3. 

2y/5  4-  307. 

2013  -  7. 
6y/2  -  305. 
14  +  20*1. 

05  +  0*. 

\/|  -  y/f. 

.    30?  -  2y/3. 

Solve  the  following  equations : 

94.    8  -  20b  =  4. 

.4ws.  4. 

95.  6  +  0b  =   2^/12  +  x. 

96.  0b  -  3  -  V^  +  12  =  -3. 

97.  ^dx  +  10  -  \/3x  +  25  +  3  =  0. 

4. 

4. 

-3. 

12. 

y/3o;  +  13 
99.    2  +  ^aj  -  5  =  13. 

fnOj.     J*  -  \[x  -  8   = ? . 

1336. 
9. 

0e  -  8 


PURE    QUADRATIC  EQUATIONS.  249 


CHAPTER     XIII. 

QUADRATIC  EQUATIONS  OF  ONE  UNKNOWN 
QUANTITY. 

130.  Quadratic  Equations.  —  An  equation  which  con- 
tains the  square  of  the  unknown  quantity,  but  no  higher 
power,  is  called  a  quadratic  equation,  or  an  equation  of  the 
second  degree. 

A  Pure  quadratic  equation  is  one  which  contains  only  the 
square  of  the  unknown  quantity ;  it  is  sometimes  called  an 
incomplete  quadratic  equation.  An  Adjected,  or  Affected? 
quadratic  equation  is  one  which  contains  both  the  square  and 
the  first  power  of  the  unknown  quantity  ;  it  is  also  called  a 
complete  quadratic  equation. 

Thus,  2a,*2  =  50,  and  ax1  +  6  =  0, 

are  pure  quadratic  equations  ;  and 

2x2  —  bx  =  4,  and  ax2  +  bx  -f-  c  =  0, 
are  affected  quadratic  equations. 

131.  Pure  Quadratic  Equations.  —  A  pure  quadratic 
may  be  solved  for  the  square  of  the  unknown  quantity  by 
the  method  of  solving  a  simple  equation. 

Let  it  be  required  to  solve 

x'2  -  13    .    x2  -  5        a 

—3—  +  ~io-  =  6* 

Clear  of  fractions, 

10.^2  -  130  -f  Sx2  -  15  =  180. 

.-.     13a;2  =  325. 

x2  =  25. 

Extracting  the  square  root  x  =  ±5. 

*  The  terra  aclfected,  or  affected,  was  introduced  by  Vieta,  about  the  year  1600,  to 
distinguish  equations  which  involve,  or  are  affected  with,  differeut  powers  of  the 
unknown  quantity  from  those  which  contain  one  power  only. 


250  PURE    QUADRATIC  EQUATIONS. 

In  this  example  we  find  that  x2  =  25.  Therefore  x  must 
be  such  a  number  that  if  multiplied  by  itself  the  product  is 
25;  i.e.,  x  must  be  the  square  root  of  25;  we  prefix  the 
double  sign  to  5  because  the  square  root  of  a  quantity  may 
be  either  positive  or  negative.     [Art.  106,  (1)]. 

Note.  —  In  extracting  the  square  root  of  the  two  members  of 
the  equation  x2  =  25,  it  might  seem  at  first  that  we  ought  to  prefix  the 
double  sign  to  the  square  root  of  each  side,  and  write  ±x  =  ±5.  An 
examination  however  of  the  various  cases  shows  this  to  be  unneces- 
sary, because  we  obtain  no  new  results  in  so  doing.  Thus,  if  we 
write  ±x  =  ±5,  we  have  the  four  cases: 

-\-x  =  +5,  +x  =  —5,  —x  =  +5,  —x  =  —5; 

but  the  last  two  are  equivalent  to  the  first  two,  and  become  identical 
with  them  on  changing  the  signs.  Hence  there  are  no  new  results 
obtained,  and  therefore  when  we  extract  the  square  root  of  the  two 
members  of  an  equation,  it  is  sufficient  to  put  the  double  sign  before 
one  member  only.     Thus  the  equation  has  two  roots,  and  only  two. 

A  pure  quadratic  equation  can  always  be  reduced  to  the 
form 

ax2  +  b  =  0 ; 

for  all  the  terms  containing  x2  may  be  reduced  to  one  term, 
as  ax2 ;  and  the  known  terms  to  another,  as  b. 

By  transposing  b,  dividing  by  a,  and  putting  q  = ,  the 

equation  may  be  written 


Such  an  equation  is  called  a  binomial  equation,  because  it 
has  but  two  terms. 

Solving  this  equation  by  extracting  the  square  root  of  each 
member,  we  have 

x  =  ±vV 

That  is,  Every  pure  quadratic  equation  has  two  roots, 
numerically  equal,  but  toith  contrary  signs. 

Hence,  for  the  solutiou  of  a  pure  quadratic  equation  we 
have  the  following 


AFFECTED    QUADRATIC   EQUATIONS.  251 

Rule. 
Find  the  value  of  the  square  of  the  unknown  quantity  by  the 
rule  for  solving  a  simple  equation,  and  then  extract  the  square 
root  of  both  members. 

EXAMPLES. 

Solve 

1.  ll.r2  -  44  =  b:c-  +  10.  Ans.  x  =   ±   3. 

2.  (x  +  2)2  =  \x  +  5.  x  =   ±    1. 

3.  — —  +  — - —  =25.  x  =   ±  .3 
1  -  2x       1  +  2x 

4.  14  -  \Jx2  -  36  =  6.  a;  =   ±10. 

132.  Affected  Quadratic  Equations.  —  An  affected 
quadratic  equation  can  always  be  reduced  to  the  form 

ax2  +  ox  +  c  =  0  ; 

for  all  the  terms  containing  x~  may  be  reduced  to  one  term, 

as  ax3 ;  those  containing  x  to  one,  as  bx ;    and  the  known 

terms  to  another,  as  c. 

b  c 

If  we  divide  bv  «,  and  put  »  =  -,  and  cr  =  — -,  the  equa- 

a  a 

tion  may  be  written 

x2  +  pas  =  g, 

where  />  and  q  are  positive  or  negative.     This  is  called  the 
General  Quadratic  Equation. 

Let  it  be  required  to  solve  this  equation.  If  the  first 
member  of  this  equation  were  a  perfect  square,  we  might 
solve  it  by  extracting  the  square  root,  as  in  Art.  131.  To 
ascertain  what  must  be  done  to  make  the  first  member  a  per- 
fect square,  let  us  compare  it  with  the  square  of  the  binomial, 

x  +  *-,  which  is  x2  -\-  px  +  —  • 
2  4 

Thus,  we  see  that  x2  -+-  px  is  rendered  a  perfect  square  by 

the  addition  of  %-  ;  i.e.,  by  the  addition  of  the  square  of  half 
4 


2-52       AFFECTED    QUADRATIC  EQUATIONS. 


rr 
the  coefficient  of  x.     Hence,  adding  A—  to  both  members,  to 


preserve  the  equality,  we  have 


x2  +  px  +  ^  =  q  +  £ 
4  4 


This  is  called  completing  the  square. 

Extracting  the  square  root  of  each  member,  we  have 


*  +  t: 


f -**£+£ 


=  -i±v/*+£ 


(i) 


Thus  there  are  too  roots  of  a  quadratic  equation. 

Note  1.  —  When  an  expression  is  a  perfect  square,  the  square  terms 
are  always  positive.  Hence,  the  coefficient  of  x1  must  be  made  +1,  if 
necessary,  before  completing  the  square. 

1.  Solve  x2  +  6x  =  27. 

Here  half  the  coefficient  of  x  is  3  ;  add  32, 
x2  +  6x  +  32  =  27  +  9  =  36. 
Extracting  the  square  root, 

x  +  3  =  ±6. 
.-.     x  =  -3  ±  6  =  3,  or  -9. 
We  may  verify  these  values  as  follows : 
Putting  3  for  x  in  the  given  equation,         9  +  18  =  27. 
Putting  —9  for  x  in  the  given  equation,  81  —  54  =  27. 
These  results  being  identical,  the  values  of  x  are  verified. 
It  will  be  well  for  the  student  thus  to  verify  his  results. 

2.  Solve  7x  =  x2  -  8. 

Transposing,  so  that  the  terms  which  involve  x  are  alone 
in  the  first  member,  and  the  coefficient  of  a;2  is  +1,  we  have 
x2  —  7x  =  8. 


AFFECTED    QUADRATIC   EQUATIONS.  253 

Here  half  the  coefficient  of  x  is  — f ; 
completing  the  square, 

x2  -  7x  -h  (J)2  =  8  +  ^  =  ¥• 

/v.    7    4.9 

.-.     a;  =  | ;  ±  f  =  8,  or  -1. 
Note  2.  — We  indicate  (I)2  in  the  first  member. 

3.  Solve  32  -  3x2  =  10a. 

Transposing,  changing  signs,  and  dividing  by  3,  so  as  to 
make  the  coefficient  of  x'2  unity  and  positive, 

O*2   -I-    10  •>•   —    32  . 

completing  the  square, 

*  +  V*  +  (I)2  =  ¥  +  ¥  =  1fL- 
.-.    *  +  f  =  ±V-. 

.-.     x*=  -J  ±  V  =  2,  or  -5f 

Note  3.  — We  add  (^)2  and  not  (if)2,  to  complete  the  square. 

4.  Solve  5a;2  +  lis  =  12. 
Dividing  by  5,  x2  +  ^-x  =  ^; 

completing  the  square, 

x2  4-  -i-i  x  4-  fliV  —  -1-2  -4-  J-2  A  —  3.61 

*     T^      5  'C    ^    VlO;'      —      5      t    ioo    -    10"0' 

«.    i     ii   _    -Lin 
.  .      x  -f  To-  —    x  ro-. 

.-.     x  =  -1}  ±  ig  =  f,  or  -3. 
Hence,  for  solving  affected  quadratic  equations,  we  have 
the 

Rule  I. 

Reduce  the  equation  so  that  the  terms  involving  the  unknown 
quantity  are  alone  in  one  member,  and  the  coefficient  of  x'2 
is  +1  ;  complete  the  square  by  adding  to  each  member  of 
the  equation  the  square  of  half  the  coefficient  of  x;  extract  the 
square  root  of  both  members,  and  solve  the  resulting  simple 
equation. 

Note  4. — There  are  other  ways  of  completing  the  square  of  an 
affected  quadratic,  which  are  convenient  in  special  cases,  and  some  of 
which  will  be  given  as  we  proceed;  but  the  method  just  explained  is 
the  most  important,  and  will  solve  every  case. 


254  AFFECTED    QUADRATIC   EQUATIONS. 

Instead  of  going  through  the  process  of  completing  the 
square  in  every  particular  example,  it  is  more  convenient  to 
apply  the  following  rule  deduced  from  formula  (1)  of  this 
Article : 

Rule  II. 

Reduce  the  equation  to  the  general  form,  x2  +  px  =  q. 
Then  the  value  of  x  is  half  the  coefficient  of  the  first  power 
of  x  tvith  a  contrary  sign,  plus  or  minus  the  square  root  of 
the  second  member  increased  by  the  square  of  half  the  same 
coefficient. 

Note  5.  —  The  student  should  use  this  method  in  practice,  and 
become  familiar  with  it,  but  at  the  same  time  be  careful  that  he  does 
not  lose  sight  of  the  complete  method. 

5.    Solve  S6x  -  Sx2  =  105. 

Transposing,  changing  signs,  and  dividing  by  3, 

x2  -  12a  =  -35. 


Therefore  by  Rule  II,        x  =  6  ±  N/-35  +  36  =  1. 
.-.     x  ==  6  ±  1 'b  7,  or  5. 

6.  Solve  Sx  ~  2  =  — 2. 

2x  —  3       x  -f  4 

c.      rf  .        Sx  -  2       Sx  -  8 

Simplifying,  =  . 

1      J     °    2x  -  3         x  -f-  4 

Clearing  of  fractions, 

3x2  +  10a;  -  8  =  Gx2  -  25x  +  24. 

Reducing,      x2  —  3¥5-.c  =  —  ^-. 

Therefore,  Rule  II,    x  =  %5  ±  V7-3;,2-  +  ifjp  =  %%1-. 

.-.     x  =  V±¥=  10§,  or  1. 

7.  Solve  x2  —  4x  =  1. 

Rule  II,  x  =  2  ±  ^1+4  =  5. 

.-.     x  =  2  ±  2.236  =  4.236,  or  -0.236. 
These  values  of   x  are    correct   only  to    three    places   of 
decimals,  and  neither  of  them  will  be  found  to  satisfy  the 
equation  exactly. 


CONDITION  FOR   EQUAL  ROOTS.  255 

If  the  numerical  values  of  the  unknown  quantity  are  not 
required,  it  is  usual  to  leave  the  roots  in  the  form 

2  +  VE,  and  2  -  V& 
8.    Solve  x2  -  10.C  =  -32. 


Rule  II,  x  =  5  ±  V-32  +  25  =  -7. 

.-.     x  =  5  ±  V^7. 

But  —7  has  no  square  root,  either  exact  or  approximate 
(Art.  106)  ;  so  that  no  real  value  of  x  can  be  found  to 
satisfy  the  given  equation.  In  such  a  case  the  quadratic 
equation  has  no  real  roots ;  the  roots  are  said  to  be  imagin- 
ary or  impossible. 

In  the  examples  hitherto  considered,  the  quadratic  equa- 
tions have  had  tivo  different  roots.  Sometimes  however, 
there  is  only  one  root.  Take,  for  example,  the  equation, 
x2  —  10a;  +  25  =  0 ;  by  extracting  the  square  root  we  have 
x  —  5  =  0  ;  therefore  x  =  5.  It  is  found  convenient  how- 
ever in  this  and  similar  cases  to  say  that  the  quadratic  has 
two  equal  roots. 

EXAMPLES. 

Solve 


9. 

x2  =  x  +  72. 

10. 

9a;  -  x'2  +  220  = 

11. 

x2  -  %x  =  32. 

12. 

V*  =  i  -  x2' 

13. 

5x2  =  8x  +  21. 

14. 

fa  +  7  =  3x  +  2 
x  —  1 

15. 

3a;  -  8       5x  -  2 

Ans 

.9, 

-  8. 

20, 

-11. 

6, 

_  16 
"3  * 

i 

-   4. 

3, 

~   h 

3, 

-    1. 

4, 

n 

2"' 

x  —  2         a;  +  5 
133.   Condition  for  Equal  Roots.  —  To  find  the  rela- 
tion that  mast  exist  between  the  known  quantities  of  a  quadratic 
equation  in  order  that  the  tivo  roots  may  be  equal* 
Take  the  general  equation 

ax2  -j-  bx  +  c  =  0. 


256  CONDITION  FOR  EQUAL  ROOTS. 


Transpose  c  and  divide  by  a, 
2    ,    b  c 


xr  H — X 
a 

Rule  II,      »  = 


_  b 
2a 

w- 

a 

^2   _ 
4a2 

V 

—  4ac 

4a2 

-b  ± 

:  \Jl/  - 

-  4ac 

-•        —  2a <P 

Denoting  the  root  corresponding  to  the  positive  surd  by 
xv  and  that  corresponding  to  the  negative  surd  by  x2,  we 
have 


-b  + 

\/b2  - 

-  4ac 

2a 

-b  - 

\lb2  - 

-  4ac 

2  2a 

Now  we  see  that  if  62  —  4ac  =  0,  these  two   roots   are 

equal,  and  each  of  them  is . 

2a 

Hence  the  relation,  b2  —  Aac  ■=  0,  is  the  condition  that  the 
two  roots  of  the  equation  ax2  -}-  bx  -f  c  =  0  may  be  equal. 

The  two  roots  are  real  and  unequal  if  62  —  4ac  is  positive, 
i.e.,  if  62  is  Algebraically  greater  than  4ac. 

The  two  roots  are  imaginary  if  b2  —  4ac  is  negative,  i.e., 
if  Z>2  is  Algebraically  less  than  4ac. 

Hence  the  two  roots  of  this  equation  are  real  and  unequal, 
equal,  or  imaginary,  according  as  b2  is  greater  than,  equal  to, 
or  less  than  4ac. 

Note  1.  —  If  either  of  the  roots  of  a  quadratic  equation  is  imaginary, 
they  are  both  imaginary. 

By  applying  these  tests,  the  nature  of  the  roots  of  any 
quadratic  may  be  determined  without  solving  the  equation. 

1.  Show  that  the  equation  2a;2  —  Gx  -|-  7  =  0  cannot  be 
satisfied  by  any  real  values  of  x. 

Here  a  =  2,  b  =  —  6,  c  =  7. 

...     b2  -  Aac  =  36  -  4  .  2  •  7  =  -20. 
Hence  the  roots  are  imaginary. 


HINDOO   METHOD    OF    COMPLETING    THE   SQUARE.     257 

Determine  the  nature  of  the  roots  of 

2.  x2  4-  ox  +  1  =  0.  Ans.  Real  and  surds. 

3.  3a;2  -  4x  -4  =  0.  Rational. 

4.  If  the  equation  x2  +  2(k  +  2)x  +  dk  =  0  has  equal 
roots,  find  k.  Ans.  k  =  4,  or  1. 

When  an  equation  is  in  the  general  form  ax2  +  bx  +  c  =  0, 
instead  of  solving  it  by  either  of  the  rules  in  Art.  136,  we 
may  make  use  of  formula  (1)  above  as  follows : 

5.  Solve  ox2  +  lis  =  12. 

Here  a  =  5,  b  =  11,  c  =  —12;  substituting  these  values 

in  (1)  

x_  -11   ±  y/(ll)^-4.5(-12) 
10 
_  -11  ±  v/361  _  -11  ±19       4  Q 

~         To         ~       To       ~  f'  OT  ~3' 

which  agrees  with  the  solution  of  Ex.  4,  (Art.  132). 
Solve  b}7  this  method  the  following : 

6.  3a;2  =  15  -  ix.  Ans.  §,  -3 

7.  2a;2  +  7jb  =  15.  ],  -5 

8.  5a;'2  +  4  +  21a;  =  0.  -4,  -I 

9.  Sx2  =  x  +  7.  1,  -J 
10.    35  4-  9a;  -  2a;2  =  0.                                             7,  -f 

Note  2.  — Though  we  can  always  find  the  roots  of  a  given  quad 
ratic  equation  by  substituting  in  formula  (1),  yet  the  student  is  advised 
to  solve  each  separate  equation  either  by  the  method  given  in  Art.  132, 
and  embodied  in  Rule  II,  or  by  one  of  the  two  following. 

134.  Hindoo  Method  of  Completing  the  Square. 

—  When  a  quadratic  equation  appears  in  the  general  form 
aar2  4-  bx  4-  c  =  0,  the  first  member  may  be  made  a  complete 
square,  without  dividing  by  the  coefficient  of  a;2,  thus  avoid- 
ing fractions,  by  another  method  (called  the  Hindoo  method), 
as  follows : 

Transpose  c,  and  multiply  by  4a, 

4a'2a;'2  4-  ±abx  =  —  4ac. 


258    HINDOO   METHOD    OF    COMPLETING    THE   SQUARE. 

Now  since  the  middle  term  of  any  trinomial  square  is 
twice  the  product  of  the  square  roots  of  the  other  two  (Art. 
41),  the  square  root  of  the  third  term  must  be  equal  to  the 
second  term  divided  by  twice  the  square  root  of  the  first 
term.  Hence,  dividing  Aabx  by  twice  the  square  root  of 
4a2#2,  i.e.,  by  4a#,  and  adding  the  square  of  the  quotient, 
b2,  to  both  members,  the  first  becomes  a  perfect  square. 
Thus, 

4a2x2  +  iabx  +  b2  =  b2  —  4ac. 

Extracting  the  square  root, 


2ax  +  b  =   ±  sjb'2  —  4«c. 

-b  ±  \lb'2  -  Aac 

.-.     x  = , 

2a 

which  are  the  same  values  we  obtained  in  (1)  of  Art.  133. 

Rule. 

Reduce  the  equation  to  the  form  ax2  +  bx  -f-  c  =  0.  Mul- 
tiply it  by  four  times  the  coefficient  of  x2 ;  add  to  each  member 
the  square  of  the  coefficient  of  x  in  the  given  equation;  extract 
the  square  root  of  both  members,  and  solve  the  resulting  simple 
equation. 

Note.  —  This  method  may  be  used  to  advantage  when  we  wish  to 
avoid  fractions  in  completing  the  square,  and  it  is  often  preferred  in 
solving  literal  equations.     (See  Note  4  of  Art.  132.) 

1.    Solve  2z2  -  bx  =  3. 

Multiply  by  four  times  2,  or  8, 

IGa;2  -  40z  =  24. 

Add  to  each  side  52,  or  25, 

16a;2  -  AOx  +  25  =  49. 

Extract  the  square  root, 

4x  -  5  =   ±7. 

.-.     x  =  5-±-Z  =  3,  or  -\. 


SOLVING   A    QUADRATIC   BY  FACTORING.  259 

Solve  by  the  Hindoo  method  the  following 


o 


2.  3a;2  +  ox  =  2.  Ana.  },  -2. 

3.  6a?  -  12  =  x.  li,  -li. 

4.  3z2  -f  2a;  =  85.  5,  -5f. 

5.  acx2  —  te  4-  acfa  =  bd.  -,  — . 

a        c 

135.  Solving  a  Quadratic  by  Factoring.  —  There  is 
still  one  method  of  solving  a  quadratic  which  is  often  shorter 
than  either  of  the  methods  already  given. 

1.  Consider  the  equation  x'2  —  2x  —  15  =  0. 
Resolving  this  into  factors  (Art.  65),  we  have 

[x  -  5)  (a;  +  3)  =  0. 

Now  it  is  clear  that  a  product  is  zero  when  any  one  of  its 
factors  is  zero ;  and  it  is  also  clear  that  no  product  can  be 
zero  unless  one  of  the  factors  is  zero.  Thus  ab  is  zero  if  a 
is  zero,  or  if  b  is  zero ;  and,  if  we  know  that  ab  is  zero,  we 
know  that  either  a  or  b  must  be  zero ;  and  so  on  for  any 
number  of  factors. 

Similarly  the  product  (x  —  5)  (x  4-  3)  is  zero,  when  either 
of  the  factors,  x  —  5,  x  +  3,  is  zero,  and  in  no  other  case. 

Hence  the  equation 

O  -  5)(x  +  3)  =  0, 
is  satisfied  if  x  —  5  ==  0,  or  if  x  +  3  =\0 ;  i.e.,  if  x  =  5, 
or  if  x  =  —3,  and  in  no  other  case. 

Therefore  the  roots  of  the  equation  are  5,  and  —3. 

2.  Solve  x2  -  5x  +  6  =  0. 
Resolving  this  into  factors,  we  have 

O  -  2)(x  -  3)  =  0. 

The  first  member  is  zero  either  when  x  —  2  =  0,  or  when 
x  —  3  =  0  ;  and  in  no  other  case.  Hence  the  equation  is 
satisfied  by  x  =  2,  or  3  ;  and  by  no  other  values  ;  thus,  2 
and  3  are  the  roots  of  the  equation. 

From  these  examples  it  appears  that  when  a  quadratic 
equation  has  been  simplified,  and  has  all  its  terms   in  the 


260  SOLVING  A    QUADRATIC  BY  FACTORING. 

first  member,  its  solution  can  alwa3's  be  readily  obtained  if 
the  expression  can  be  resolved  into  factors.  Hence  for  the 
solution  of  such  an  equation,  we  have  the  following 

Rule. 

Reduce  the  equation  to  its  simplest  form,  with  all  its  terms 
in  the  first  member;  then  resolve  the  whole  expression  into 
factors,  and  the  values  obtained  by  equating  each  of  these 
factors  separately  to  zero  will  be  the  required  roots. 

3.  Solve  x2  —  4x  =  0. 

Factoring,  we  have  x(x  —  4)  =  0. 

The  equation  is  satisfied  if  x  =  0,  or  if  x  —  4  =  0,  and 
in  no  other  case.     Hence  we  must  have  either 

x  =  0,  or  x  —  4  =  0. 

.-.     x  =  0,  or  4. 

Note  1.  —  In  this  example  we  might  have  divided  the  given 
equation  by  x  and  obtained  the  simple  equation  x  —  4  =  0,  whence 
x  =  4,  which  is  one  of  the  solutions.  But  the  student  must  be  parties 
ularly  careful  to  notice  that  whenever  we  divide  every  term  of  an 
equation  by  x,  it  must  not  be  neglected,  since  the  equation  is  satisfied 
by  x  z=  0,  which  is  therefore  one  of  the  roots. 

Note  2. — When  the  factors  can  be  written  down  by  inspection, 
the  student  should  always  solve  the  example  in  this  way,  as  he  will 
thus  save  himself  a  great  deal  of  unnecessary  work. 

Solve  the  following  by  resolution  into  factors  : 

4.  (3x  -  l)(3x  +  1)  =  0.  Ans.  ±J. 

5.  x2  -  lias  =  0.  0,  11. 

6.  aj2  —  Bx  +  2  ==  0.  1,  2. 

(%  x2  -  2x  =  8.  4,  -2. 

8.  or2  -  lax  +  4a6  =  2bx.  2a,  26. 

9.  x2  -  2ax  +  8x  =  ICa.  2a,  -8. 

Note  3.  —  When  the  student  cannot  factor  the  equation  readily  by 
inspection,  he  should  solve  it  by  Rule  11,  Art.  lo2,  or  by  Art.  lo4. 


TO  FORM  A  QUADRATIC   WHEN  THE  ROOTS  ARE  GIVEN.    2G1 

136.  To  Form  a  Quadratic  when  the  Roots  are 
Given.  —  We  have  seen  (Art.  135)  that  if 

x2  +  px  -\-  q  =  {x  —  a)  (x  —  b), 
then  a  and  b  are  the  roots  of  the  equation 

x*  +px  +  q  =  0 (1) 

Conversely,  if  a  and  b  are  roots  of  (1),  then  x  —  a  and 
x  —  b  are  factors  of  the  expression  x2  -J-  px  -f-  q,  which 
ma}7  be  proved  as  follows  : 

Since  a  is  a  root  of   (1),  we  have 

a2  +  pa  +  q  =  0 (2) 

Hence  x2  -f  Vx  +  </  =  ^2  +  p#  +  <?  —  («2  +  i*^  +  <?) 
=  (»  —  a)  (x  +  a  +  2?) . 
.-.     cc  —  a  is  a  factor  of  x2  +  jpaj  +  ^. 
Hence,  if  a  is   a   root   of  (1),  x  —  a  will   be   a   factor  of 
x2  +  1KB  +  q- 

Similarly  it  may  be  shown  that  if  b  is  a  root  of  (1),  then 
x  —  b  will  be  a  factor  of  x2  +  px  +  g. 

Now  SB2  -\-  px  -\-  q  cannot  have  more  than  two  factors  of 
the  form  x  —  a,  for  the  product  of  any  number  of  factors 
of  the  form  x  —  a  must  be  of  the  same  degree  in  x  as  the 
number  of  the  factors ;  also  x2  -\-  px  -\-  q  clearly  has  no 
factors  not  containing  x. 

Hence  x2  -(-  px  -+-  q  =  (x  —  a)  (x  —  b)     .     .     .    (3) 

Performing  the  multiplication  in  (3),  we  have 
x2  -f-  px  +  q  —  x2  —  (a  +  b)x  +  ab. 
Hence  we  have      a  +  6  =  —  p  \  ,*\ 


ab  =       q. 

That  is,  wi  a  quadratic  equation  where  the  coefficient  of  x2  is 
unity  and  all  the  terms  are  in  the  first  member,  the  sum  of  the 
roots  is  equal  to  the  coefficient  of  x  with  its  sign  changed,  and 
the  product  of  the  roots  is  equal  to  the  third  term. 

Thus,  dividing  the  general  equation  by  a,  it  becomes 

x2  +  h  +  ±  =  0  .     .     .     .     .     .   (5) 

a         a 


262    TO  FORM  A  QUADRATIC  WHEN  THE  ROOTS  ARE  G/VEN. 

Adding  together  the  two  roots  of  (1),  Art.  133,  we  have 
x^  -\-  x2  —  ; 

and  by  multiplication  we  have 

4a2  a 

which  confirms  the  proposition. 

Hence  any  quadratic  may  be  expressed  in  the  form 

x2  —  (sum  of  roots)  x  +  product  of  roots  =  0   .    (6) 

By  this  principle  we  may  easily  form  a  quadratic  with 
given  roots.  Although  we  cannot  in  all  cases  find  the  roots 
of  a  given  equation,  it  is  easy  to  solve  the  converse  problem, 
namely,  the  problem  of  finding  an  equation  which  has  given 
roots. 

These  relations  are  useful  in  verifying  the  solution  of  a 
quadratic  equation.  If  the  roots  obtained  do  not  satisfy 
these  relations,  we  know  there  is  some  error  in  the  work. 

Relations  analogous  to  those  above  hold  good  for  equations 
of  the  third  and  of  higher  degrees.  But  we  defer  the  proof 
to  a  subsequent  chapter. 

When  we  know  one  root  of  an  equation,  we  may  by 
division  lower  the  degree  of  the  equation.  Thus  if  a  is  one 
root  of  an  equation,  we  may  divide  it  by  the  factor  x  —  a. 

Note  1.  — In  any  equation  the  term  which  does  not  contain  the 
unknown  quantity  is  frequently  called  the  absolute  term. 

A  quadratic  equation  cannot  have  more  than  two  roots.  For 
no  other  value  of  x  besides  a  and  b  can  make  (as  —  a)(x  —  b) 
in  (3)  equal  to  0,  since  the  product  of  two  factors  cannot 
equal  0  if  neither  factor  is  equal  to  0. 

It  may  therefore  be  inferred  that  a  cubic  equation  has  three 
and  only  three  roots ;  and  that  in  any  equation  the  number  of 
roots  is  equal  to  the  degree  of  the  equation. 

Note  2.  —  The  student  must  carefully  distinguish  between  a  quad- 
ratic equation  and  a  quadratic  expression.  Thus,  iu  the  quadratic 
equation  x-  -\-  px  +  q  =  0,  we  must  suppose  x  to  have  one  of  two 


EQUATIONS   HAVING   IMAGINARY  ROOTS.  263 

definite  values,  or  roots,  but  when  we  speak  of  the  quadratic  expres- 
sion x2  +  px  -f-  q,  without  saying  that  it  is  to  be  equal  to  zero,  we  may 
suppose  x  to  have  any  value  we  please. 

EXAMPLES. 

Form  the  quadratic  equation  whose  roots  are 

1.  2  and  3.  Am.  x2  —  5x  +  6  =  0. 

2.  3  and  -2.  a*  -  x  -  6  =  0. 
3..  2  +  S/3  and  2  -  V^.                            x2  -  Ax  +1  =  0. 

4.  6  and  8.  x2  -  Ux  +  48  =  0. 

5.  4  and  5.  x2  -  9x  +  20  =  0. 
137.   Equations  Having  Imaginary  Roots.  —  It  was 

shown  (Art.  133)  that  when  b2  is  less  than  4ac,  i.e.,  when 

b2  c 

— -  is  less  than  -,  the  two  roots  are  imaginary.  Hence,  from 
4a2  a 

(5)  and  (6)  of  Art.   136,  the  roots  are  imaginary  ivhen  the 

square  of  half  their  sum  is  less  than  their  product.     Now  it 

is  impossible  to  have  two  numbers  such  that  the  square  of 

half  their  sum   is  less  than  their  product,   which  may  be 

shown  as  follows : 

Let  a  represent  any  number ;  and  suppose  it  to  be  divided 

into  two  parts  -  +  x  and  -  —  x.     Then  the  product  is 

a2         2 

r  -  *  ' 

which  is  evidently  the  greatest  when  x2  is  the   least.     But 

when  x2  or  x  =  0  the  parts  are  each  - ;  thus  the  product  of 

two  unequal  numbers  is  less  than  the  square  of  half  their 
sum.     Hence, 

The  square  of  half  the  sum  of  tivo  numbers  can  never  be 
less  than  their  product. 

If  then  any  problem  furnishes  an  equation  of  the  general 
quadratic  form  xM  +  px  +  q  =  0,  in  which  q  is  positive  and 

greater  than  the  square  of  *-,  we  infer  that  the  conditions  of 


264   EQUATIONS  OF  HIGHER  DEGREE  THAN  THE  SECOND. 

the  problem  are  incompatible  with  each  other,  and  hence  the 
problem  is  impossible.     Thus, 

Let  it  be  required  to  divide  6  into  two  parts  whose  product 
shall  be  10. 

Let  x  =  one  of  the  parts, 
then  6  —  x    =  the  other. 

.-.     a(6  —  a;)  =  10, 
whence  x  =  3±V—  1. 

Thus,  the  roots  are  imaginary.  Now  we  know  from  the 
preceding  proposition,  that  the  number  6  cannot  be  divided 
into  any  two  parts  whose  product  will  be  greater  than  9. 
Hence  when  we  are  required  to  divide  6  into  two  parts  whose 
product  is  10,  we  are  required  to  solve  an  impossible  problem. 
Thus,  the  imaginary  root  shows  that  the  problem  is  impossible. 

138.  Equations  of  Higher  Degree  than  the  Second. 

—  There  are  many  equations  which  are  not  really  quadratic, 
but  which  may  be  reduced  to  the  quadratic  form,  and  solved 
by  the  methods  explained  in  this  Chapter.  An  equation  is 
in  the  quadratic  form  when  the  unknown  quantity  is  found  in 
two  terms,  and  its  exponent  in  one  term  is  twice  as  great  as 
in  the  other.     Thus, 

x*  -  9x2  =  -20,  (x2  +  x)2  +  i(x2  +  x)  =  12, 
ax2n  +  bxn  +  c  =  0,  etc., 

are  in  the  quadratic  form,  and  may  be  solved  by  either  of 
the  preceding  rules ;  care  however  should  be  taken  to  use 
the  one  best  adapted  to  the  example  considered. 

1.   Solvere4  -  9x2  =  -20. 

Here  we  may  complete  the  square  and  solve  by  Rule  I, 
Art.  132,  or  we  may  write  out  the  value  of  x2  at  once  by 
Rule  II,  Art.  132  as  follows : 


x2  =  §  ±  V7-  20  +  V  =  i 

=  \  ±  \  =  5,  or  4. 
x  =   ±V/;">,  or  ±2. 
Thus  there  are  four  roots,  ±V5,  ±2. 


EQUATIONS  OF  HIGHER  DEGREE  Til  AX  THE  SECOND.    265 

Otherwise  thus. 

Transposing  and  factoring  the  first  member, 
(a:2  -  5)  (a2  -  4)  =  0. 
.•.     x2  —  o  =  0,  giving  x  =   ±Vo, 
or  x'2  —  4  =  0,  giving  a;  =   ±2. 

2.  Solve  as*  +  6a?"  +  c  =  0. 
Transpose  and  divide  by  a, 

x2n  +  -x»  =  -c-. 
a  a 


Art.  132,  Rule  II,  xn  =  -—  ±J --  +  — 

2a      V       a       4tr 


b2  —  4\ac 
4a2 


-5  ±  S/b2  -  4ac 
2a 
from  which  x  may  be  found  by  taking  the  nth  root  of  both 
members. 

Note  1.  —If  the  student  prefers,  he  may  let  xn  =  y\  then  x2n  =  y2. 
Substituting,  the  equation  becomes 

oy2  +  by  +  c  =  0. 
After  solving  for  y,  he  may  replace  the  value  of  y. 

3.  Solve   x*  -  6.r3  =  16. 

Art.  132,  Rule  II,  Xs  =  3  ±^2h  =  3  ±  5  =  8,  or  -2. 
.-.     se  =  2,  or  —  y2. 

4.  Solve  a;-*  +  a:-5  =  6. 


Solving  for  aT±,  x~*  =  —  |  ±  V/6  +  J  =  2,  or  -3. 

.-.     a;-1  =  1G.  or  81. 

•••     ■*      =  A>  of  A- 

5.    Solve  vV  +  12  +  Vx2  -j-  12  =  6. 
Solving  for  yar  +12, 

fo2  +  12  =  -\  ±  s/6  +  i  =  -i  ±  f  =  2,  or  -3. 
.-.     x2  +  12  =  16,  or  81. 
.-.     x2  =  4,  or  69. 

.-.     a*  =   ±2,  or  ±^69. 


266    EQUATIONS  OF  HIGHER  DEGREE  THAN   THE  SECOND. 


6.  Solve  x  -f  \/dx  +10  =  8. 
By  transposing  x,  and  squaring, 

bx  4-  10  =  64  -  16a;  +  x\ 
.-.     x2  -  21a;  =  -54. 
Solving,  we  get  x  =  -^  ±  ^  =  18,  or  3. 

If  we  proceed  to  verify  these  values  of  x  by  substituting 
them  in  the  given  equation,  we  shall  find  that  3  satisfies 
the  equation,  but  that  18  does  not,  while  it  does  satisfy  the 
equation 

x  —  sjbx  +  10  =  8. 
Now  the  reason  is  this  :  the  equation  x2  —  21a;  =  —54, 
which  we  obtained  from  the  given  equation  by  transposing 
and  squaring,  might  have  been  obtained  as  well  from 
x  —  \5x  +  10  =  8,  since  the  square  root  of  a  quantity 
may  have  either  the  sign  -|-  or  —  prefixed  to  it;  i.e.,  the 
resulting  equation  x2  —  21a;  =  —54,  of  which  18  and  3  are 
the  roots,  would  be  obtained,  whether  the  sign  of  the  radical 
be  4-  or  — .  Hence  we  see  that  when  an  equation  has  been 
reduced  to  a  rational  form  by  squaring,  we  cannot  be  certain 
without  trial  whether  the  values  which  are  finally  obtained 
for  the  unknown  quantity  are  roots  of  the  given  equation. 

7.  Solve  x2  -  7x  +  \Jx2  -  7z  +  18  =  24. 

Add  18  to  both  members  in  order  that  the  equation  may  be 

in  the  quadratic  form. 

x2  -  7x  +  18  -f-  V^2  -  7a;  +  18  =  42. 


Solving,  \x2  -  7a;  4-18  =  -J  ±V42  +  J 

=  —J-  ±  ^  =  6,  or  -7. 
.-.     x2  -  7a;  4-  18  =  36,  or  49. 
Solving  the  first  quadratic,  we  obtain      a;  =  9,  or  —2. 
Solving  the  second  quadratic,  we  obtain  x  =  -J (7  ±  Vl73). 
Only  the  first  two  values  are  roots  of  the  given  equation  ; 
the  other  two  are  roots  of  the  equation 

x2  -  7x  -  Var  -  7a;  -f  18  =  24. 


SOLUTIONS   BY   FACTORING.  207 

8.    Solve  x4  -  4.r3  -  2a-2  +  12a;  -  10  =     0. 

We  proceed  to  form  a  perfect  trinomial  square  with  the 
first  two  terms  and  a  part  of  the  third.  The  square  root  of 
this  square  is  evidently  (ar  —  2x),  the  square  of  which  is 
x*  —  4x?  +  4x2 ;  having  added  4ar  to  the  equation  we  must 
now  subtract  4x2.     Hence  the  equation  becomes 

xA  -  4a;3  +  4ar  -  Gx2  +  12a;  -  10  =     0, 
or  (a?  -  2x)2  -  6(x2  -  2x)  =  10, 

which  is  in  the  quadratic  form,  and  may  be  solved  as  those 
above. 

Hence  x2  —  2x  =  3  ±  VlG  +  9  =  8,  or  —  2. 


.-.     x  =  1  ±  ^8  +  1,  or  1  ±  yJ-2  +  1. 
.*.     x  =  4,  or  —2,  or  1  ±  v— 1. 
Solve  the  following : 

9.    x4  -  13a-2  +  30  =  0.  Ans.  ±2,  ±3. 

10.  x2  +  V^2  +  9  =  21.  ±4. 

11.  9y/x*  -  9a;  +  28  -f-  9a;  =  x2  +  36.  12,  -3. 

12.  of"  -f-  ^  =  1050.  64,  (-33)*. 

13.  (a;2  -  9)2  -  ll(a?  -  2)  =  3.  ±5,  ±2. 

14.  x4  -  8x*  +  10a?  -f  24a;  -f  5  =  0.  5,  -1,  2  ±  \[b. 

139.  Solutions  by  Factoring.  —  By  the  principles  of 
Art.  135,  many  equations  of  a  higher  degree  than  the  second 
may  be  solved,  which  cannot  be  reduced  to  the  quadratic 
form.     If  an  equation  can  be  reduced  to  the  form 

(x  -  a)X  =  0, 

in  which  X  represents  an  expression  involving  a;,  we  have 

either  A  v       n 

x  —  a  =  0,    or  A  =  0  ; 

therefore  x  =  a,  is  one  value  of  x ;  and  if  we  solve  the 
equation  X  =  0,  we  shall  have  the  other  values  of  x.  Hence 
whenever  we  have  one  factor  of  an  equation,  we  have  at 
least  one  root,  and  by  division  we  may  lower  the  degree  of 
the  equation  by  one  (Art.  130).     Thus 


268  SOLUTIONS   BY   FACTORING. 

1.  Solve  (x  -  h){x2  -  Sx  4  2)  =  0. 

Here  the  first  member  is  zero  either  when  x  —  5  =  0,  or 

when  x2  —  3x  -f  2  =  0 ;  and  in  no  other  case.     Hence  we 

have 

x  -  5  =  0,  or  x2  -  Sx  +  2  =  0. 

From  the  first  we  have  aj  =  5  ;  and  the  other  roots  of  the 
equation  are  those  given  by 

a;2  —  Sx  +  2  =  0, 

that  is,  (a?  -  2)  (a;  -  1)  =  0. 

Thus  the  cubic  equation  (x  —  5)  (ic2  —  3#  +  2)  =  0  has 
the  three  roots  5,  2,  and  1. 

The  difficulty  to  be  overcome  in  this  method  consists  in 
resolving  the  equation  into  factors ;  and  facility  in  separat- 
ing expressions  into  factors  can  be  acquired  only  by  experi- 
ence. 

2.  Solve  xs  -  1  =  0. 

Since  xs  -  1  =  (aj  -  l)(x2  +  x  -f-  1), 

we  have  (x  —  1)  {x2  -f  x  +  1)  =  0. 

...     x  -  1  =  0,  or  x2  +  x  +  1  =  0, 

the  roots  of  which  are  1,  or  —J  ±  V— f. 

Hence  there  are  three  roots  of  the  equation  x3  =  1,  one 
being  real  and  the  other  two  imaginary.  Thus  there  are 
three  numbers  whose  cubes  are  equal  to  1  ;  that  is,  there 
are  three  cube  roots  of  1. 

3.  Solve   x  —  1  =  2  4-  ~.  Ans.  4. 

Sx 

4.  "       2a;3  -  x2  -  Gx  =  0.  0,  2,  -f. 

5.  "      x3  +  x2  -  4x  =  4.  -1,  2,  -2. 
G.       "      a;3  -  3aJ  =  2.  -1,  2. 

7-    -    *2-32,=  1»-  -fKi±Vio); 


PROBLEMS   LEADING    TO    QUADRATIC    EQUATIONS.    269 

140.  Problems  Leading  to  Quadratic  Equations 
of  One  Unknown  Quantity.  —  We  shall  now  give  some 
examples  of  problems  which  lead  to  pure  or  affected  quad- 
ratic equations  of  one  unknown  quantity. 

In  the  solution  of  such  problems,  the  equations  are  found 
on  the  same  principles  as  in  problems  producing  simple 
equations  (Art.  61). 

EXAMPLES. 

1.  Find  two  numbers  such  that  their  sum  is  15,  and  their 
product  is  54. 

Let  x  =  one  of  the  numbers, 

then  15  —  x  =  the  other  number. 

Hence  from  the  conditions,  we  have 
#(15  —  x)  =  54, 
or  x2  —  15a;  =  —54. 

Solving,  we  sjet  x  =  9,  or  6. 

If  we  take  x  =  9  we  have  15  —  x  =  6 ;  and  if  we  take 
x  =  6  we  have  15  —  x  =  9.  Thus,  whichever  value  of  x 
we  take,  we  get  for  the  two  numbers  6  and  9.  Hence, 
although  the  equation  gives  two  values  of  x,  yet  there  is 
really  only  one  solution  of  the  problem. 

2.  A  man  buys  a  horse  which  he  sells  again  for  896  ;  he 
finds  that  he  thus  loses  one-fourth  as  much  per  ceut  as  the 
horse  cost  him  :  find  the  price  of  the  horse. 

Let    x  =  the  price  of  the  horse  in  dollars, 

xl 
then  - —  =  the  man's  loss  in  dollars. 
400 

Hence  from  the  conditions,  we  have 

x2 

JL    =  x  -  96. 
400 

Solving,  we  get  x  —  240  or  160. 

That  is.  the  price  was  either  $240  or  Si  60,  for  each  of 
these  values  satisfies  the  conditions  of  the  problem. 


270  EXAMPLES. 

3.  A  train  travels  300  miles  at  a  uniform  rate  ;  if  the  rate 
had  been  5  miles  an  hour  more,  the  journey  would  have  taken 
two  hours  less  :  find  the  rate  of  the  train. 

Let  x  =  the  rate  the  train  runs  in  miles  per 

hour ; 
then  300  -r-  x  =  the  time  of  running  on  the  first  sup- 

position ; 
and  300  -i-  (x  -+-  5)  =  the  time  of  running  on  the  second 
supposition. 

300  300      0 

—  — . 


x  -f  5 
Solving,  we  get  x  =  25,  or  —30. 

Only  the  positive  value  of  x  is  admissible,  and  thus  the 
train  runs  25  miles  per  hour. 

Note.  —  In  the  solutions  of  problems  it  often  happens  that  the 
roots  of  the  equation,  which  is  the  Algebraic  statement  of  the  relation 
between  the  magnitudes  of  the  known  and  unknown  quantities,  do 
not  all  satisfy  the  conditions  of  the  problem.  The  reason  of  this  is 
that  the  Algebraic  statement  is  more  general  than  ordinary  language; 
and  the  equation,  which  is  a  proper  representation  of  the  conditions, 
will  also  express  other  conditions.  Thus,  the  roots  of  the  equation 
are  the  numbers,  whether  positive,  negative,  integral,  or  fractional, 
which  satisfy  that  equation ;  but  in  the  problem  there  may  be  restric- 
tions on  the  numbers,  expressed  or  implied,  which  cannot  be  retained 
in  the  equation.  If  for  instance,  one  of  the  roots  of  an  equation  is  a 
fraction,  it  cannot  be  a  solution  of  a  problem  which  refers  to  a  number 
of  men,  for  such  a  number  must  be  integral.     Thus 

4.  Eleven  times  the  number  of  men  in  a  group  is  greater 
by  twelve  than  twice  the  square  of  the  number :  find  the 
number  of  men  in  the  group. 

Let  x  =  the  number  of  men  ;  then  we  have 
11a;  =  2x2  +  12, 
or  2x2  -  11a;  =  -12. 

Solving,  we  get  x  =  4,  or  1£. 

Thus,  there  are  4  men  ;  the  value  1J  is  plainly  inadmissi- 
ble. 


EXAMPLES.  271 

5.  Eleven  times  the  number  of  feet  in  the  length  of  a  rod 
is  greater  by  twelve  than  twice  the  square  of  the  number  of 
feet :  how  long  is  the  rod  ? 

This  question  leads  to  the  same  equation  as  Ex.  4,  only 
here  we  cannot  reject  the  fractional  result,  since  the  rod  may 
be  either  4  feet  long  or  H  feet  long. 

6.  The  square  of  the  number  of  dollars  a  man  possesses 
is  greater  by  600  than  ten  times  the  number :  how  much  has 
the  man? 

Let  x  =  the  number  of  dollars  the  man  has. 
Then  x*  =  10a:  +  600. 

Solving,  we  get  x   =  30,  or  —20. 

Both  these  values  are  admissible,  since  a  negative  posses- 
sion is  a  debt  (Art.  20). 

7.  The  sum  of  the  ages  of  a  father  and  his  son  is  80  years  ; 
also  one-fourth  of  the  product  of  their  ages,  in  years, 
exceeds  the  father's  age  by  240  :  how  old  are  they? 

Let  x  =  the  father's  age  in  years ; 

then  80  —  x  =  the  son's  age  in  years. 

Hence  Jz(80  -  x)  =  x  +  240, 
or  x2  —  76x  =  —960. 

.*.     x  =  60,  or  16. 

Thus  the  father  is  60  and  the  son  20  years  old. 

The  second  solution  16  is  not  admissible,  since  it  would 
make  the  father  younger  than  his  son. 

Note. — The  student  should  examine  each  root  of  every  equation 
to  see  if  it  satisfies  the  conditions  of  the  problem,  and  reject  those 
which  do  not. 

8.  A  cistern  can  be  filled  by  two  pipes  in  33J  minutes ;  if 
the  larger  pipe  takes  15  minutes  less  than  the  smaller  to  fill 
the  cistern,  find  in  what  time  it  will  be  filled  by  each  pipe 
singly.  Ans.  75  and  60  minutes. 

9.  A  person  selling  a  horse  for  S72,  finds  that  his  loss  per 
cent  is  one-eighth  of  the  number  of  dollars  that  he  paid  for 
the  horse  :  what  was  the  cost  price?  Ans.  §80,  or  §720. 


272  EXAMPLES   OF    QUADRATICS. 

10.  Divide  the  number  10  into  two  parts  such  that  their 
product  added  to  the  sum  of  their  squares  may  make  76. 

Ans.  4,  6. 

11.  Find  the  number  which  added  to  its  square  root  will 
make  210.  Ans.  196. 

12.  A  and  B  together  can  do  a  piece  of  work  in  14f  days  ; 
and  A  alone  can  do  it  in  12  clays  less  than  B  alone :  find  the 
time  in  which  A  alone  can  do  the  work.  Ans.  24  days. 

13.  A  company  dining  together  at  an  inn,  find  their  bill 
amounts  to  $35  ;  two  of  them  were  not  allowed  to  pay,  and 
the  rest  found  that  their  shares  amounted  to  $2  a  man  more 
than  if  all  had  paid  :  find  the  number  of  men  in  the  company. 

Ans.  7. 

14.  The  side  of  a  square  is  110  inches  long  :  find  the  length 
and  breadth  of  a  rectangle  which  shall  have  its  perimeter 
4  inches  longer  than  that  of  the  square,  and  its  area  4  square 
inches  less  than  that  of  the  square.  Ans.  126,  96. 

Note.  —  We  will  conclude  this  chapter  with  the  following  examples. 
In  solving  them  care  must  be  taken  to  select  the  method  best  adapted 
to  the  example  considered.  Many  of  them  may  be  solved  by  special 
methods  (Arts.  133,  134,  135);  but  the  methods  of  Art.  132  are  the 
most  important,  and  will  solve  every  example. 

EXAMPLES    OF    QUADRATICS. 


1.   xy(j  +  x2  =  1  +  x2.  Ans.  ±J, 


7a?a  +  8         x2  +  4 


21  8a;2  -  11       3 


±2. 


1                                          1                          X  — 

4'  *  +  <jt=*  +  <»--va--*  =  *'  ±v/3" 

5.  x2  —  5a;  =  —6.  2,  3. 

6.  x2  —  10a,*  =  -9.  1,  9. 

7.  x2  +  12a;  =  13.  1,  -13. 


EXAMPLES 

273 

8. 

x2  +  9a;  =  22. 

Ans.  2,  -11. 

9. 

x2  +  2a;  =  143. 

11,  -13. 

0. 

15ar  —  2aa;  =  a2. 

a        a 
3'        5' 

11.  21a;2  =  2ax  +  3a*.  fa,  -* 

12.  JLtJl  =  2JL^J.  4,  # 


2a;  —  7        x  —  3 
1  1 


19. 

x  +  1        a;  —  2  _  9 
tT  _  1       x  _j_  2        5' 

20. 

1                 2       _  3 

a;  —  2       a;  +  2        5 

21. 

3                       1 

2(a;2  -  1)        4(a;  +  1)        8' 

22. 

5            3  _        14 
x  4-  2       a;       a;  +  4 

23. 

4                5               12 
a;  +  1       x  +  2  ~  a;  +  3* 

24. 

a;  +  1        a;  —  1  _  2a;  —  1 

a;  -  2 


j  j- 


1  _|_  x       3  -  x       ^'  iU'  ~3' 

14.    — —  =  1  3^  _, 

a;  —  1        x  -\-  2        x 


15.    (a  +  \){2x  +  3)  =  4a;2  -  22.  5,  -2J. 

4-  a;2  _  x  — 
3  2 


16.    2  +  ^  _  ^zJ:2  =  1  -  a;  +  a?.  1,  2. 


17.  8a;  +  11  +  I  =  — .  7,  -&. 

a;  7  " 

18.  3(*  ~  *>  -  ^L±J)  =  5i  |,  _3. 

x  +  1  a;  -  1  3 

3,-f 

3,  -4§. 

3,  -5. 

3,  -1|. 


4,0. 


25.    ?Ll^  +  :L±^  =  2(*L±J\  H,0. 

a;  +  2       a;  -  2         \a;  -  3/ 


274  EXAMPLES. 


Qn     x  —  1    .    x  —  2        2x  +  13  A        -        n  . 

26. =  — Ans.  5,  —  1A. 

x  +  1        a;  +  2        a;  +  16  ° 

w-  3L+1  +  -  +  «  s  gg  +  13.  5  lh 

x  -  1       a?  —  2         a;  +  1  '    5 

28.  2^~1  +  8*~1  =  5*  ~  U.  5,  -U. 
x  +  1         a;  +  2          a;—  1 

14r  —  9        r2  —  3 

29.  a;  -  — = 2«,  0. 

8a;  -  3         a;  4-  1  3 

,1       i    i    1 
a?  +  -       1  +  - 

30.   5  + 2  =  8*.  S,  -If. 

a; 1 

a;  a? 

si.  v^+_2  =  *-_y*  4. 

4  +  y/a;  y/a; 

32.  oV  -  2a3a;  +  a4  -  1  =  0.  a  ±-. 

a 

33.  4a2x  =  (a2  -  b2  +  x)2.  (a  ±  b)2. 

34-   -  +  -  =  -  +  -  ±v/a6. 

a       a;       6       a;  v 

35.  (3a2  +  b2){x2  -  x  +  1)  =  (362  +  a2)  (a2  +  a>  +  1). 

a  —  b     a  +  b 

36.  —|  =1  +  1  +  1.  _«,  -6. 

a  +  o  +  x       a       b       x 

37.  a  +  c(a  4-  a?)       a  +  «?  _.        g  _a    «(1  4-  c) 
a  4-  c(a  —  x)           x          a  —  2ca;  '  c(2c  +  3) 

x    ,    a       2 


38.  =  +  =  -  -  =  0.  1  ±  y/i  _a« 

•7.  01*.  f I.  » 


39.    -—5 — F- (a* -&*)»=  -* -•   a>  ~&- 


EXAMPLES.  275 

Form  the  equation  whose  roots  are 

40.  1  ±  5.  Ans.  x2  -  2x  -  A  =  0. 

41.  -£,  f.  35ar  +  13a;  -  12  =  0. 

42-  f-^|  -J^|-    (p2  -  s2)*2  +  *pq*  -  vl  +  g2  =  o. 

43.  7  ±  2y/5.  a;2  -  14a;  4-  29  =  0. 

44.  ±2y73  -  5.  a2  +  10a;  +  13  =  0. 

45.  -p  ±  2y/2g.  x2  +  Spg  +  p2  -  8g  =  0. 

Show  that  the  roots  of  the  following  equations  are  real : 

46.  x2  -  2ax  +  a2  -  b2  -  c2  =  0. 

47.  (a  —  b  4-  c)a;2  +  4 (a  —  b)x  +  (a  —  b  -  c)  =  0. 

For  what  values  of  m  will  the  following  equations  have 
equal  roots? 

48.  x2  -  15  -  m(2x  —  8)  =  0.  -4ws.  3,  5. 

49.  x2  -  2£(1  +  3m)  4-  7(3  4-  2m)  =  0.  2,  — V0-. 

EXAMPLES    OF    EQUATIONS    REDUCIBLE    TO 
QUADRATICS. 

50.  a;4  -  5a;2  +  4  =  0.  Ans.  ±1,  ±2. 

51.  (a;2  -  a;)2  -  8(a;2  -  a;)  -f  12  =  0.  3,  -1,  ±2. 


.    -1   ±  sJ-27 


52.    (a;2  +  x)  (x2  +  x  +  1)  =  42.     2,  -3, 


53. 


^  +  IJ+4^  +  ^=12.  1,-3±2V^. 

54.  a;2  4-  3  -  ^2x2  -  3x  +  2  =  §(<c  4-  1).  1,  i- 

55.  x4  4-  2a;3  -  11a;2  +  4a;  4-  4  =  0.  1,2. 

56.  x4  4-  4afa  =  a4.  --£  ±  ^2^2_-  1§ 

y/2  ^2 

(i  +  xy     * 


4ns.  1  4-  y/3  ±  y/3  +  2y3>  1  -  V  ^  ±  V3  -  203. 


276 


EXAMPLES. 

841 
3ar 

+ 

J  7 
3 

_232 
.     3a; 

Ans. 

1 
"  3a;2 

2,- 

58.  27x2  -  ^=  +  -V-  =  ™  -^+5. 

l|,i(-2  ±V^266). 

Solve  as  explained  in  Art.  143  : 

59.  2a;3-a;2=l.  Ans.  1,.  J(-1±V^7). 

60.  a;3  -6a;  =  9. 

61.  8a;3  +  16a;  =  9. 

62.  a;3  -  3a;2  +  x  +  2  =  0. 

63.  3a;6  +  8a;4 -8a?  =  3.  ±1 

64.  a;3  +  a;2  -  4a;  -  4  =  0. 

65.  a;3  +  aa;2  +(a  -  1  +  — ^-  V  +  1 
a—  1/ 


3,*( 

-3±V-3). 

ii(- 

-1  ±  V-35). 

2,  1(1  ±  S/5). 

.[i(- 

-11  ±  ^86)]*. 

-  1,  ±  2. 

=  0. 

VS 


l-a,H-l  ± 

66.  Find  a  number  whose  square  diminished  by  119  is 
equal  to  ten  times  the  excess  of  the  number  over  8.  Ans.  13. 

67.  A  man  is  five  times  as  old  as  his  son,  and  the  sum  of 
the  squares  of  their  ages  is  equal  to  2106  :  find  their  ages. 

Ans.  45,  9. 

68.  If  a  train  traveled  5  miles  an  hour  faster  it  would 
take  one  hour  less  to  travel  210  miles :  what  time  does  it 
take?  Ans.  7  hours. 

69.  The  sum  of  the  reciprocals  of  two  consecutive  numbers 
is  |f  :  find  them.  Ans.  7,  8. 

70.  The  perimeter  of  a  rectangular  field  is  500  yards,  and 
its  area  is  14400  square  yards :  find  the  length  of  the  sides. 

Ans.  90  and  160  yards. 

71.  A  number  of  two  digits  is  equal  to  twice  the  product 
of  the  digits,  and  the  digit  in  the  ten's  place  is  less  by  3 
than  the  digit  in  the  unit's  place :  what  is  the  number? 

Ans.  36. 

72.  The  sum  of  a  certain  number  and  its  square  root  is 
42:  what  is  the  number?  Ans.  36. 


EXAMPLES.  211 

73.  A  rectangular  court  is  ten  yards  longer  than  it  is 
broad;  its  area  is  1131  square  yards:  find  its  length  and 
breadth.  Ans.  39  and  29. 

74.  One  hundred  and  ten  bushels  of  coal  were  divided 
among  a  certain  number  of  persons ;  if  each  person  had 
received  one  bushel  more  he  would  have  received  as  many 
bushels  as  there  were  persons  :  find  the  number  of  persons. 

Ans.  11. 

75.  A  cistern  can  be  supplied  with  water  by  two  pipes  -, 
by  one  of  them  it  would  be  filled  6  hours  sooner  than  by  the 
other,  and  by  both  together  in  4  hours :  find  the  time  in 
which  each  pipe  alone  would  fill  it.  Ans.  6,  12. 

76.  Two  messengers  A  and  B  were  sent  at  the  same  time 
to  a  place  90  miles  distant ;  the  former  by  riding  one  mile 
per  hour  more  than  the  latter  arrived  at  the  end  of  his 
journey  one  hour  before  him  :  find  at  what  rate  per  hour 
each  traveled.  Ans.  10,  9  miles. 

77.  A  person  rents  a  certain  number  of  acres  of  land  for 
$280 ;  he  keeps  8  acres  in  his  own  possession,  and  sublets 
the  remainder  at  $1  per  acre  more  than  he  gave,  and  thus  he 
covers  his  rent  and  has  88  over :  find  the  number  of  acres. 

Ans.  56. 

78.  From  two  places  320  miles  apart,  two  persons  A  and 
B  set  out  in  order  to  meet  each  other.  A  traveled  8  miles  a 
day  more  than  B  ;  and  the  number  of  days  in  which  they 
met  was  equal  to  half  the  number  of  miles  B  went  in  a  day : 
find  how  far  each  traveled  before  they  met.     Ans.  192,  128. 

79.  A  certain  number  is  formed  by  the  product  of  three  con- 
secutive numbers,  and  if  it  be  divided  by  each  of  them  in  turn, 
the  sum  of  the  quotients  is  47  :  find  the  number.       Ans.  60. 

80.  A  boat's  crew  row  2>h  miles  down  a  river  and  back 
again  in  1  hour  and  40  minutes  :  supposing  the  river  to  have 
a  current  of  2  miles  per  hour,  find  the  rate  at  which  the  crew 
would  row  in  still  water.  Ans.  5  miles  per  hour. 

81.  A  person  rents  a  certain  number  of  acres  of  land 
for  $336  ;  he  cultivates  4  acres  himself,  and  letting  the  rest  for 


278  EXAMPLES. 

$2  an  acre  more  than  lie  pays  for  it,  receives  for  this  portion 
the  whole  rent,  $336  :  find  the  number  of  acres.  Ans.  28  acres. 

82.  A  person  bought  a  certain  number  of  sheep  for  &140  ; 
after  losing  two  of  them  he  sold  the  rest  at  $2  a  head  more 
thau  he  paid  for  them,  and  by  so  doing  gained  $4  by  the 
transaction:  find  the  number  of  sheep  he  bought.      Ans.  14. 

83.  A  man  sends  a  lad  to  the  market  to  buy  12  cents' 
worth  of  oranges  ;  the  lad  having  eaten  a  couple,  the  man 
pays  at  the  rate  of  one  cent  for  15  more  than  the  market 
price  :  how  many  did  the  man  get  for  his  12  cents?    Ans.  18. 

84.  A  person  drew  a  quantity  of  wine  from  a  full  vessel 
which  held  81  gallons,  and  then  filled  up  the  vessel  with 
water ;  he  then  drew  from  the  mixture  as  much  as  he  before 
drew  of  pure  wine ;  and  it  was  found  that  G4  gallons  of  pure 
wine  remained :  find  how  much  he  drew  each  time. 

Ans.  9  gallons. 

85.  A  certain  company  of  soldiers  can  be  formed  into  a 
solid  square ;  a  battalion  consisting  of  seven  such  equal 
companies  can  be  formed  into  a  hollow  square,  the  men 
being  four  deep.  The  hollow  square  formed  by  the  battalion 
is  sixteen  times  as  large  as  the  solid  square  formed  by  one 
company  :  find  the  number  of  men  in  the  company.    Ans.  64. 

86.  Find  that  number  whose  square  added  to  its  cube  is 
nine  times  the  next  higher  number.  Ans.  3. 

87.  A  courier  proceeds  from  one  place  P  to  another  place 
Q  in  14  hours  ;  a  second  courier  starts  at  the  same  time  as 
the  first  from  a  place  10  miles  behind  P,  and  arrives  at  Q 
at  the  same  time  as  the  first  courier.  The  second  courier 
finds  that  he  takes  half  an  hour  less  than  the  first  to  go  20 
miles  :  find  the  distance  from  P  to  Q.  Ans.  70  miles. 

88.  A  vessel  can  be  filled  with  water  by  two  pipes ;    by 
one  of  these  pipes  alone  the  vessel  would  be  filled  2  hours 
sooner  than  by  the  other;    also  the  vessel  can  be  filled  by' 
both  pipes  together  in  1J  hours:    find  the  time  which  each 
pipe  alone  would  take  to  fill  the  vessel.    Ans.  5  and  3  hours. 


SIMULTANEOUS    QUADRAT  I  ('   EQUATIONS.  279 


CHAPTER    XIV. 

SIMULTANEOUS     QUADRATIC     EQUATIONS. 

141.  Simultaneous  Quadratic  Equations. — We  shall 
now  consider  some  of  the  most  useful  methods  of  solving 
simultaneous  equations,  one  or  more  of  which  may  be  of  a 
degree  higher  than  the  first.  It  should  be  remarked  that  it 
is,  in  general,  impossible  to  solve  a  pair  of  simultaneous 
quadratic  equations  ;  for,  if  we  eliminate  one  of  the  unknown 
quantities,  the  resulting  equation  will  be  of  the  fourth  degree 
with  respect  to  the  other  unknown  quantity,  and  we  cannot,  in 
general,  solve  an  equation  of  a  higher  degree  than  the  second. 

There  are  several  cases  however  in  which  a  solution  of 
two  equations  may  be  effected,  when  one  or  both  of  them 
are  of  the  second  or  some  higher  degree.  Various  artifices 
are  employed  for  the  solution  of  such  equations,  the  proper 
application  of  which  must  be  learned  by  experience. 

142.  Case  I.  When  One  of  the  Equations  is  of 
the  First  Degree.  —  This  case  may  be  solved  by  the 
following 

Rule.  From  the  simple  equation  find  the  value  of  one  of 
the  unknown  quantities  in  terms  of  the  other,  and  substitute 
this  value  in  the  other  equation. 

1.    Solve  3a;  +  4y  =  18 (1) 

5x>  -  3xy  =     2 (2) 

From  (1)  we  have  y  = '-  ; (3) 

4 

and  substituting  in  (2), 

&*  -  3*<18  ~  3*>  =  2. 
4 

.-.     20a?  -  54a;  +  9s2  =  8, 

or  29a?  —  54a;  =  8. 


280  SIMULTANEOUS    QUADRATIC   EQUATIONS. 

Solving,  we  get  x  =  2,     or     —  2%; 

and  by  substituting  in  (3),  y  =  3,     or     -2^6g7. 
Solve  the  following : 

2.  3a;   —  4?/  =  5,                         Arts,  x  =  3,  or  —if1, 
3a;2  -  xy  -  Sy2  =21.                      ?/  =  1,  or  -  A£. 

3.  5x  —     ?/  =  17,                                 a;  =  4,  or  —  f, 

a#  =12.                                 y  =  3,  or  -  20. 

4.  2<e  —  5y  =    3,                                 x  =  4,  or  —  -2T5-, 

a;2  +  ^  =  20.                                 ?/=  1,  or  -|f 

5.  4a;  —  5?/  =     1,                                  a;  =  4,  or  —  f§, 
2a;2  -  a#  +  3?/2  +  3a;  -  Ay  =  47.      2/  =  3>  or  -  ti- 

143.  Case  II.    Equations  of  the  Form  x  ±  y  =  at 
and  xy  =  b ;  or  jr2  +  /2  =  a,  and  xy  —  b. 

1.   Solve                                 a?  +  y  =    15  .  .     .  .  (1) 

ay  =    36  ....   (2) 

Square  (1),                 x2  +  2oy  +  2/2  =  225 ;  .     .  .   (3) 

multiply  (2)  by  4,                            Axy  =  144 ;  .     .  .   (4) 
subtract  (4)  from  (3),  x2  -  2xy  +  y2  =    81 ; 

extract  the  square  root,              x  —  y  =   ±9.  .     .  .   (5) 

Combining  (5)  with  (1),  we  have  the  two  cases 


x  +  y  =  15,)      x  +  y  =     15,) 
x  -  y  =    9.)      x  -  y  =  -9. J 


from  which  we  have  a?=12,)  »  =      3,) 

2/=    3.1  2/=     12.  J 

2.   Solve  a2  +  ?/2=  25 (1) 

*W  =  12 (2) 

Multiply  (2)  by  2  ;  then  by  addition  and  subtraction  we  have 

a;2  +  2a7/  +  f  =     49, 

a;2  —  2xy  +  y2  =       1. 

.-.     x  +  ?/  =   ±7, 

x  —  y  =  ±1. 


SIMULTANEOUS    QUADRATIC    EQUATIONS.  281 

We  have  now  four  cases  to  consider,  namely, 
x  +  y=7)     x  +  y=      1\     x  +  y=  —  7)     <c  +  y=-7) 
x  —  y  =  1  )     .T-?/=-lf     x-y  =      1  J      a;  —  y  =  —  1 J 

From  which  the  values  of  x  are  4,  3,  —3,  —  4  ; 

and  the  corresponding  values  of  y  are  3,  4,  —4,  —3. 

Thus  there  are  four  pairs  of  values,  two  of  which  are  given 
by  x  =  ±4,  y  =  ±3,  and  the  other  two  by  x  =  ±3,  ?/  =  ±4, 
it  being  understood  that  in  both  cases  the  upper  signs  are  to 
be  taken  together,  and  the  lower  signs  are  to  be  taken 
together. 

These  are  the  simplest  forms  that  occur,  but  the}'  are 
specially  important,  since  the  solution  of  a  large  number  of 
other  forms  is  dependent  upon  them.  Our  object,  as  a  rule, 
is  to  solve  the  given  equations  symmetrically,  by  finding  the 
values  of  x  -f-  y  and  x  —  y,  which  we  can  always  do  as  soon 
as  we  have  obtained  the  product  of  the  unknown  quantities 
and  either  their  sum  or  their  difference. 

Any  pair  of  equations  of  the  form 

x-  ±  pxy  +  y2  =  a2 (1) 

x  ±  y  =  b (2) 

where  p  is  any  numerical  quantity,  can  be  reduced  to  one 
of  the  forms  above  considered  ;  for  by  squaring  (2)  and 
combining  with  (1),  we  obtain  an  equation  to  find  xy  \  the 
solution  can  then  be  completed  by  the  aid  of  equation  (2). 
Thus 

3.    Solve  x2  +  xy  +  y2  =  333 (1) 

x-y  =  3 (2) 

Square  (2),         x2  -  2xy  -f-  y2  =  9 (3) 

Subtract  (3)  from  (1),  Sxy  =  324. 

.-.     xy  t=  108 (4) 

Add  (1)  and  (4)  and  extract  square  root, 

x  +  y  =   ±21 (5) 

From  (2)  and  (5)  x  =  12,  or  —9. 

y  =  9,  or  -12. 


282  SIMULTANEOUS   QUADRATIC  EQUATIONS. 

4.   Solve  -  -  -  =  i (1) 

x       y 

h  +  h  =  i (2> 

x2       y2 

Square  (1),  \  -  i-  +  \  =  J (3) 

x2       xy       y 

Subtract  (3)  from  (2),      —  =  | (4) 

Add  (2)  and  (4)  and  take  the  square  root, 

l  +  i=  ±1 (5) 

a;       ?/ 

From  (1)  and  (5),       -  =  |,  or  -J. 


as 

3,  or  -f. 


1  _  i 


2/ 

.-.     x  =  |,  or  —3,  and  ?/  =  3,  or  — f . 

Solve  the  following : 

5.  x  —  y  =  12,                             Ans.  x  =  17,  or  —  5, 

xy  =  85.                                        y  =     5,  or  —17. 

6.  re  +  y  =  74,                                       &  =  53,  or      21, 

xy  =  1113.                                   y  =  21,  or      53. 

7.  a  +  ?/  =  84,                                        a;  ==  71,  or      13, 

xy  =  923.                                     y  =  13,  or      71. 

8.  x2  +  ?/2  =  97>                                    a;  =    9,  or        4, 
a    +  ?/   =  13.                                     ?/  =    4,  or        9. 

9.  x  +  ?/  =  9,                            a;  =    5,  or        4, 
x2  +  xy  +  y2  =  61.                          y  =    4,  or        5. 

144.  Case  III.    When  the  Two  Equations   Con- 
tain a  Common  Algebraic  Factor. 

Rule.    Divide  one  equation  by  the  other,  and  cancel  the 
common  factor. 


SIMULTANEOUS    QUADRATIC   EQUATIONS.  283 

1.    Solve  coy  +  y-  =  133 (1) 

V2  =     95 (2) 


.7' 


2  „2 


Divide  (2)  by  (1)  and  cancel  the  common  factor  x  +  y, 


x  ~  V  _    95     _    5 
y      ~  **  ~  7* 


X  = 


12 


iftri (3) 


and  substituting  in  (1) 

iff  +  y2  =  133. 
Solving,  we  get  y  =  ±7 ; 

and  by  substituting  in  (3),      cc  =  ±12. 

Note.  —  This  includes  the  case  where  one  equation  is  exactly 
divisible  by  the  other. 

2.  Solve  x?  +  ys=18xy.     .     .     .     .  (1) 

x  +  y  =  12 (2) 

Divide  (1)  by  (2),    x2  -  xy  +  y2  =  fa#. 

•••     x*-%xy  +  f  =  0 (3) 

Subtract  (3)  from  the  square  of   (2), 

%xy  =  144. 

.'.     i^y  =16 (4) 

Add  (3)  and  (4)  and  take  the  square  root, 

x  -  y  =  ±4 (5) 

From  (2)  and  (5)  x  =  8,  or  4, 

?/  =  4,  or  8. 

3.  Solve  x4  +  ajy  +  ?/4  =  2613 (1) 

x2  +  xy  +  if  =       07 (2) 

Divide  (1)  by  (2),   x2  -  xy  +  y2  =      30 (3) 

Add  (2)  and  (3),  x2  +  ?/2  =       53. 

Subtract  (3)  from  (2),  a#   =       14. 

.*.     x  =  ±  7,     or     ±2, 
y   =   ±   2,     or     ±7. 


284  SIMULTANEOUS    QUADRATIC  EQUATIONS. 


Solve  the  following : 

4.    x'2  -f-  xy  =  84, 

Ans. 

,  #  = 

±7, 

x2  - 

y2  =  24. 

y  = 

±5. 

5.    x4  + 

x2y2  +  ?/4  = 

133, 

#  = 

±3,  or  ±   2, 

x2  + 

^   +  2/2  = 

19. 

2/  = 

±2,  or  ±   3. 

6.    a;3  + 

2/3  =  407, 

#  = 

7,  or        4, 

x    + 

2/  =     11- 

2/  = 

4,  or        7. 

7.x    — 

2/  =      4, 

x  = 

11,  or  -    7, 

x3  - 

?/3  =  988. 

2/  = 

7,  or  -11. 

145.  Case  IV.  When  the  Two  Equations  are  of 
the  Second  Degree  and  Homogeneous. 

First  method. 

1.    Solve  2x2+    3xy  +      y1  =  70    .     .     .     .   (1) 

6x2  +       xy  -       if  =  50    .     .     .     .   (2) 

Divide  (1)  by  (2),  2f+    ^  +      ^  =  J. 

Hence  \0x2  +  loxy  +    hy2  =  42z2  +  7a#  -  7y2; 

or  32z2  -     8xy  -  12y2  =  0. 

This  is  a  quadratic  equation  which  we  may  solve,  and  find 
the  value  of  one  unknown  quantity  in  terms  of  the  other. 
Thus,  solving  for  #, 

x2  -  \xy  =  |?/2. 


By  Art.  132,  Rule  H,  x  =  |  ±  y/|/  +  £  =  ff  */2, 
or  #  =  2 — - — 2.  —  j^  or  —  £?/. 

b 

Take  a;  =  £?/,  and  substitute  in  either  (1)  or  (2),  and  wc 
have  y2  =  16. 

.-.     ?/  =  ±4,     and     a?  =  ±3. 

If  we  substitute  the  second  value  of  x,  which  is  —  \y,  we 
find  an  inadmissible  result. 

Second  method.  Examples  of  this  class  are  conveniently 
solved  by  substituting  for  one  of  the  unknown  quantities  the 


SIMULTANEOUS    QUADRATIC  EQUATIONS.  28") 

product  of  the  other  and  a  third  unknown  quantity,  called 
an  auxiliary  quantity. 

2.    Solve             2y2  -  Axy  +  Sx2  =  17 (1) 

y2  -    x2  =  16 (2) 

Put  y  =  wc,  and  substitute  in  (1)  and  (2).     Thus 

x2(2v2  -  4v  +  3)  =  17 (3) 

x2(  v2  -              1)  =  16 (4) 

By  division, 


2v2  -4^  +  3      _  i7 
v2  -  1  "  ** 


.•.     32-u2  -  64v  +  48    =  17v2  -  17; 
or  I5v2  -  64v  =  -G5. 

Solving,  we  obtain  v  =  f ,  or  -^. 

Take  v  =  f,  and  substitute  in  either  (3)  or  (4), 
From  (4)      x2{^  -  1)  =  16;         .-.     x2  =  9. 

.*.     x  =  ±3,     and    y  =  «a?  =  f aj  =  ±5. 
Again,  take  v  =  V,  «^(W    —  1)  =  16  ;         .-.     x2  =  ^. 
.*.     x  =  ±f,     and     y  =  vx  =  ±1£. 
Any  pair  of  equations  which  are  o/  £/ie  second  degree  and 
homogeneous,   can  be  solved   by  either  of   these  methods, 
though  the  second  is  usually  preferred. 
Solve  the  following : 

3.  x2  +  3xy  =  28,  Ans.  x  =  ±4,  or  if  14, 
a$    +  4y2  =8.  y  =  ±1,  or  ±  4. 

4.  a2  +    a#  +  2/  =74,  x  =  ±8,  or  ±   3, 
2x2  +  2a#  -f    ?/2  =  73.                       y  =  ^5,  or  ±   5. 

5.  x2  +    a?y—    6?/2  =  24,  a;  =  ±6, 
a;2  +  3xy  -  10y2  =  32.                                    y  =  ±2. 

6.  a2  +    a#  -  6/  =21,  a;  =  ±9, 

#?/  —  2y2  =4.  ?/  =  ±4. 

Note.  —  Many  examples  of  homogeneous  equations  of  the  second 
degree  are  easily  solved  by  Case  II  or  III.  Only  those  examples  of 
this  class  are  to  be  solved  by  Case  IV  that  cannot  be  solved  by  either 
Case  II  or  III. 


286  SYMMETRICAL   EQUATIONS. 

146.  Case  V.  When  the  Two  Equations  are 
Symmetrical  with  respect  to  x  and  /.  —  An  expres- 
sion is  said  to  be  symmetrical  with  respect  to  two  letters  when 
these  letters  are  similarly  involved,  i.e.,  when  they  can  be 
interchanged  without  altering  the  expression.  Thus,  the 
expression  a3  +  a2x  +  «^2  -f-  #3  is  symmetrical  with  respect 
to  a  and  x,  since  if  we  write  x  for  a,  and  a  for  x,  we  get  the 
same  expression.  Also  a;4  +  3x2y  -f  3xy2  +  ?/4  is  symmet- 
rical with  respect  to  x  and  y. 

Examples  of  this  class  may  frequently  be  solved  by 
substituting  for  the  unknown  quantities,  the  sum  and  differ- 
ence of  two  others. 

1.    Solve  a4  +  y*  =  82 (1) 

x  -  y  -    2     .     .    .     :     I    .  (2) 

Put  x  =  u  -+-  v,     and     y  =  u  —  v, 

(2)  becomes        (u  -f  v)  —  {u  —  v)  =     2  ;         ,-.     v  =  1. 
(1)  becomes     (w  +  l)4  +  (u  -  l)4  =  82. 
.-.     2(w4  +  6it2  +  1)  =  82; 
or  u*  +  6u2  -  40  =     0. 

Hence,  Art.  135,  (u2  -f-  10)  (tt2  -  4)  =     0. 


.*. 

u    =  ±2,  or  ±\/-10. 

Thus 

a;    =  3,  -1,      1  ±  \/-10, 

2/    =  1,-3,  -1  ±  V-10. 

2.    Solve 

(x2  +  2/2)(^3  +  2/3)  =  280     ...     .   (1) 

x  +  ?/  =       4     ....   (2) 

Put 

a;  =  ?«  +  v,     and     y  =  u  —  v\ 

(2)  becomes 

(u  +  v)  +  (w  —  v)  =  4  ;         .-.     m  =  2. 

Also     x2  -f- 

?/2  =  (2  +  <y)2  +  (2  -  v)2  =      8  +     2v2, 

and  x3  +  ?/3  =  (2  +  ^)3  +  (2  -  '^)3  =     16  +  12*>2. 


SPECIAL    METHODS.  28^ 

Hence  (1)  becomes  (8  +  2V2)  (16  +  12v2)  =  280, 
or  (4  +  V2)(4  +  36-)  =     35. 

...       tf   +   J^LyS   =       n, 

.'.      v2  =  -I  ±  ^=  1,   or -If. 


0  =  ±1,  or  ±v/--139-- 


.-.     a  =      3,  1,  2  ±  y/-1^ 
y  =      1,3,  2  T  y/31^. 
Solve  the  following : 

3.  a?  —  y  =  2,     and     a;5  —  y5  =  242. 

^4ns.  a;  =  3,  or  —  1  ;     ?/  =  1,  or  —3. 

4.  x  —  ?/  =  1,     and     a5  —  if  =  781. 

.4tts.  £  =  4,  —3  ;     y  =  3,  —4. 

5.  a;  -f  y  =  3,     and     a5  +  y5  =  33. 

Ans.  x  =  1,  2;     y  =  2,  1. 

147.  Special  Methods.  —  The  preceding  cases  will  be 
sufficient  as  a  general  explanation  of  the  methods  to  be 
employed  ;  but  in  some  cases  special  artifices  are  necessary. 
One  that  is  often  used  with  advantage  consists  in  consider- 
ing the  sum,  difference,  product,  or  quotient,  of  the  two 
unknown  quantities  as  a  single  quantity,  and  first  finding  its 
value.  Other  artifices  may  also  be  used  with  advantage,  but 
familiarity  with  them  can  be  obtained  only  by  experience. 

1.    Solve   x2  +  Axy  +  3x  =  40  -  6y  -  Ay2    .     .     .    (1) 

2xy  -  x2  =  3 (2) 

From  (1)  we  have  x2  +  Axy  +  4y2  +  3x  +  Gy  =  40  ; 

or  (x  +  2y)2  +  S(x  +  2y)  -  40  =  0. 

Consider  x  ■+■  2y  as  a  single  unknown  quantity,  and  find 
its  value  from  this  quadratic.     Thus, 

(Art.  135),  [(a  +  2y)  +  8] [(a  +  2y)  -  5]  =  0. 

.-.    x  +  2y  =  -8, (3) 

or  x  +  2y  =      5    v.     •    .     .     .  (4) 


288  SPECIAL   METHODS. 

From  (2)  and  (4)  we  obtain 

x  =  1 ,  or  f ; 

y  =  2,  or  J. 
From  (2)  and  (3)  we  obtain 

x  =  ~4  ±  V15, 

2 

-12  T  y/To 

*  2 

2.  Solve  x*y*  -  6jb  =  34  -  Sy (1) 

Bxy  +  V  =  18  +  2x (2) 

Multiply  (2)  by  3  and  subtract  the  result  from  (1), 
x2y2  -  9xy  -f  20  =  0. 
.-.     (xy  -  5)(xy  -  4)  =  0. 

.-.    xy  =  5    .     .   v.    .  (3) 
xy  =  4:    .     .     .     .  (4) 
From  (2)  and  (3)  we  obtain 

x  =  1,  or  -f, 
?/  =  5,  or  -2. 
From  (2)  and  (4)  we  obtain 

x  =  ~3  ±  V^17,      and      ?/  =  3  ±  \/l7. 

Solve  the  following : 

3.  x2  +  y=73  -3x-2xy,     Ans,  x  =  4,  10,  -12  ±  ^58, 
y2+x=U-3y.  2/=5,  -7,  -1  q:  ^bS. 


4. 

^  +  **  =  12, 

2/      y 

x  —  y  =  3. 

x  =  6,    V, 
2/  =  3,  -». 

5. 

x2  H-  3a;?/  =  54, 

a;  =  ±3,  ±36. 

6. 

xy  -|-  4^2  =  115. 

a4  -  «2  +  y4  -  y2 

=  84, 

2/=    ±5,   T^. 

x  =   ±3,  ±2. 

x2  +  x2y2  +  2T 

=  49. 

y  =  ±2,  ±3. 

PROBLEMS   LEADING    TO    QUADRATIC   EQUATIONS.    289 

148.  Simultaneous  Quadratic  Equations  with 
Three  Unknown  Quantities. 

1.  Solve  xy  +  xz  =  27 (1) 

yz  +  yx  =  32 (2) 

gS   -+-  3#  =  35 (3) 

Add  (1)  and  (2)  and  subtract  (3)  from  the  sum, 

2xy  =  24;         .-.     xy  =  12  .     .     .     .   (4) 

Subtract  (4)  from  (1),      a»  =     15 (5) 

Subtract  (4)  from  (2),      yz  =     20 (G) 

Multiply  (4)  and  (5),    x2yz  =180 (7) 

Divide  (7)  by  (G),  x2  =       9       .♦.     x  =  ±3. 

Hence  from  (4),  y  =     12  h-  (±3)  =   ±4. 

And  from  (5),  z  =     15  -*-  (±3)  =  ±5. 

Thus  x  =  ±3,  y  =  ±4,  2  =  ±5, 

all  the  upper  signs  being  taken  together. 
Solve  the  following : 

2.  3yz  -h  22KB  —  4txy  =  16,  Ans.  x  =  ±1, 
2#3  -  3za  -f  a#  =5,  ?/  =  ±2, 
4?/z  —     z.r  —  Sxy  =  15.                                      2  =    ±3. 

3.  6(3?  +  2/2  +  z2)  =  13(a;  +  y  +  z)  =  ±§±,     x  =  f ,  |, 

a#  =  z2-  2/  =  h  li 

2   =    ±2. 

149.  Problems  Leading  to  Simultaneous  Quad- 
ratic Equations. 

1.  The  small  wheel  of  a  bicycle  makes  135  revolutions 
more  than  the  large  wheel  in  a  distance  of  2G0  yards ;  if  the 
circumference  of  each  were  one  foot  more,  the  small  wheel 
would  make  27  revolutions  more  than  the  large  wheel  in  a 
distance  of  70  yards  :  find  the  circumference  of  each  wheel. 

Let  x  =  the  circumference  of  the  small  wheel  in  feet, 
and      y  =  the  circumference  of  the  large  wheel  in  feet. 

Then  the  two  wheels  make  —  and   —   revolutions  respec- 

x  y 

tively  in  a  distance  of  2GU  yards. 


200  PROBLEMS   LEADING   TO    QUADRATIC  EQUATIONS. 

Hence 


780 

_ 

780 

— 

135; 

X 

y 

1 

X 

_  i 
y 

= 

9 
52       ' 

Similarly  from  the  second  condition, 
210  210 


a:  +  1        2/  +  1 


=  27; 


or  — L- ?--  =  79o (2) 

aj  +  1       y  +  1       70  V 

t^         /■•  \  52?/  ...        61?/  +  52 

From(l),£=- ^— -;       .*.     a; -f  1  = — lZ—[-— -. 

w  9?/  +  52  9y  +  52 

Substituting  in  (2),   9y  +  52 ?L_  -    9 

°        V   ^'61?/  +  52       y  +  1        70 

.-.     9?/2  -  113?/  =  52. 

Solving,  we  obtain        ?/  =  13,  or  —  f . 

Substituting  y  =  13  in  (1),  we  find  x  =  4.  The  negative 
value  of  y  is  inadmissible. 

Hence  the  small  wheel  is  4  feet,  and  the  large  wheel  is  13 
feet  in  circumference. 

2.  A  man  starts  from  the  foot  of  a  mountain  to  walk  to 
its  summit ;  his  rate  of  walking  during  the  second  half  of  the 
distance  is  half  a  mile  per  hour  less  than  his  rate  during 
the  first  half,  and  he  reaches  the  summit  in  5^  hours.  He 
descends  in  3|  hours  by  walking  at  a  uniform  rate,  which  is 
one  mile  per  hour  more  than  his  rate  during  the  first  half  of 
the  ascent :  find  the  distance  to  the  summit,  and  the  rates 
of  walking. 

Let  2x  =  the  number  of  miles  to  the  summit, 

and        y  =  the  rate  of  walking,  in  miles  per  hour,  during 

the  first  half  of  the  ascent. 

x 
Then    -  =  the  time  in  hours  for  the  first  half  of  the  ascent ; 

y 

and =  the  time  in  hours  for  the  second  half  of  the 

y-i 

ascent. 


PROBLEMS   LEADING   TO    QUADRATIC   EQUATIONS.    291 

Hence  -+  -£—  =  5J (1) 

Similarly  — ?^-  =  3f (2) 

From  (2)  x  =  *$(y  +  1) (3) 

From  (1)       x(2y  -  J)  =  V^/0/  -4) (4) 

Substituting  (3)  in  (4), 

¥(y+  i)(2z/  -  i)  =  Vz/(2/  -  i). 

.-.     28y2  -  89y  =  -15. 
Solving,  we  obtain  y  =  3,  or  55F. 

Substituting  ?/  =  3  in  (3),  we  find  a;  =  i^5.  The  other 
value  of  y  is  inapplicable,  because  by  supposition  y  is  greater 
than  J. 

Hence  the  whole  distance  to  the  summit  is  15  miles,  and 
the  rates  of  walking  are  3,  2J,  and  4  miles  per  hour. 

3.  The  sum  of  the  squares  of  two  numbers  is  170,  and 
the  difference  of  their  squares  is  72  :  find  the  numbers. 

Ans.  11  ;  7. 

4.  The  product  of  two  numbers  is  108,  and  their  sum  is 
twice  their  difference  :  find  the  numbers.  Ans.  G  ;  18. 

5.  The  product  of  two  numbers  is  G  times  their  sum,  and 
the  sum  of  their  squares  is  325  :  find  the  numbers. 

Ans.  10  ;  15. 

6.  A  certain  rectangle  contains  300  square  feet ;  a  second 
rectangle  is  8  feet  shorter,  and  10  feet  broader,  and  also 
contains  300  square  feet :  find  the  length  and  breadth  of  the 
first  rectangle.  Ans.  20  ;   15. 

7.  Find  two  numbers  such  that  their  sum  may  be  39,  and 
the  sum  of  their  cubes  17199.  Ans.  15  and  24. 

8.  The  product  of  two  numbers  is  750,  and  the  quotient 
of  one  divided  by  the  other  is  3^ :  find  the  numbers. 

Ans.  50  and  15, 


292  EXAMPLES. 

EXAMPLES    OF    SIMULTANEOUS    QUADRATICS. 

Note.  —  In  the  great  variety  of  simultaneous  quadratic  equations, 
it  is  impossible  to  give  rules  for  every  solution.  The  artifices  employed 
in  Algebraic  work  are  very  numerous.  The  student  is  cautioned  not 
to  go  to  work  upon  a  pair  of  equations  at  random,  but  to  study  them 
until  he  sees  how  they  can  be  reduced  to  a  simpler  equation  by 
addition,  multiplication,  factoring,  or  by  some  other  process,  and  then 
to  perform  the  operations  thus  suggested. 

Solve  the  following : 

1.  x  +  Ay  =  14, 
if  +  Ax  =  2y "+  11. 

2.  Sx  +  2y  =  16, 

xy  =  10. 

3.  x    +  2y    =  9, 
Sy  -  5ar  =  43. 

4.  3a;  —    y  =  11, 

3ar  -if  =  47.- 

5.  Ax   4-  9?/   =  12, 
2a;2  +  xy  =     Gy'2. 

6.  3a;  +  2y  =  5a;?/, 
15a;  —  Ay  =  Axy. 

7.  x   +     y  =     51, 

xy  =  518. 

8.  x  —    y  =       18, 

xy  =  1075. 

9.  x2  +    y2  =  89, 

xy   =  40. 

10.  x2+    y2=  178, 
x  +    y  =     16. 

11.  a;2  +    ?/2=     185, 
»  -     ?/  =         3. 


Ans.  x  = 

2, 

-46, 

2/  = 

3, 

15. 

x  = 

2, 

10 

2/  = 

5, 

3. 

a;  = 

1, 

—  11 

17' 

2/  = 

4, 

112 
17   ' 

a;  = 

4, 

7, 

y  = 

1, 

10. 

x  = 

-24, 

i. 

y  = 

12, 

4 

a;  = 

f ' 

o, 

2/  = 

3 

0. 

x  = 

37, 

14, 

2/  = 

14, 

37. 

a;  = 

43, 

-25, 

2/  = 

25, 

-43. 

a;  = 

±   8, 

±   5, 

V  = 

±    5, 

±   8. 

.T  = 

13, 

3, 

2/  = 

3, 

13. 

a;  = 

11, 

-   8, 

y  = 

8, 

-11. 

10, 

4, 

4, 

10. 

7, 

-  4/ 

4, 

-   7. 

h 

3 

8' 

1, 

1- 

EXAMPLES.  293 

12.  B3  -  xy  -f  2/2  =  76,                   <4ws.  as  = 
aj  +  y  =  14.  ?/  = 

13.  x  —  y  =    3,  a;  = 
aJ2  -  3a#  +  2/-  =  -19.  y  = 

11  J.  481  n.    

14'      -,   +  f  ~    57*>  *  - 

£  +  ! ;  =  M-  2/  = 

15«    \+h=  96oV,  *  =  ±  6,  ±  5, 
a;2       y2 

xy   =30.  y  =  ±   5,  ±   6. 

16.  x2  +  ?/2  +  a#  =  208,  a;  =  12,  4, 

a?  4-  y  =    16.  y  =  4,  12. 

17.  a-2  -  /-  =16,  a;  =  5, 
a;    -  y   =         2.  2/  =  3. 

18.  a-3  —  ?/3  =         7ajy,  a?  =  4,  —  2, 
a-    -  y    =         2.  2/  =  2,  -   4. 

19.  a-    +  y   =       23,  a?  =  14,  9, 
x3  +  y3  =  3473.  y  =  9,  14. 

20.  x    +  y   =      35,  a;  =  8,  27, 
a?*  +  y*  =        5.  ?/  =  27,  8. 

21.  x    -y   =sjx  +  V£,  a-  =  16,  9, 
ajl  -  yi  =  37.  y  =  9,  16. 

22.  a4  +  %Y  +  2/4  =  2923,  se  =  ±   7,  ±   3, 
a;2  —  xy     4-  ?/  =37.  y  =  ±   3,  ±    7. 

23.  a-4  +  a-y2  4-  2/4  =  9211,  x  =  ±   9,  ±   5, 
a-2  -  a-?/     +  y2  =61.  y  =  ±  5,  ±   9. 

24.  a:3  -  t/3  =  56,  a;  =  4,-2, 
^  +  xy  +  tf  =  28.  2/  =  2,  -  4. 

25.  a;3  4-  y3  =  126,  a;  =  5,  1, 
SP2  -  a?  4-  r  =  21.  2/  =  1,  5. 


294  EXAMPLES. 


26. 

1     ,    1  _  i    i 

~3    +    -3   -    X  TTO 

Ans.  x  =        5,        1, 

1+i-lf 

2/  =         1,        5. 

«        2/ 

27. 

a;2  -|-  a;?/  =  45,  y2  +  xy  =  36. 

a*=±   5;  ?/=±4. 

28. 

2a;2  -  a*?/  =  56,  2a*?/  -  y2  =  48. 

*=±    7  ;  2/=  ±6. 

29. 

a*2  —  2a*?/  =15,  a'?/  —  2y2  =  7. 

a*=±15;  y=±7. 

30. 

a*2#(a*+?/)=80,a;2?/(2a*-3?/)=80.       ar=±   4;?/=±l. 

31. 

«*    +    1    =    Cty, 

X    A    "2>              1) 

x2  +  x  =  Gy. 

y  =  i,  i,     o. 

Note.  —  It  will  be  seen  that  Examples  17  to  31  can  be  solved  by 

ise  III. 

32. 

x'1  +  3a*?/  =  54, 

^4ns.  jb  =  ±3,  ±36, 

xy  +  4?/2  =  115. 

y  =  ±5,  +  %3. 

33. 

x2  -\-  xy  =  24, 

«  =  ±4,  ±GV^2, 

2?/2  +  3xy  =  32. 

y  =  ±2,  T8\^2. 

34. 

x2  -  3xy  =  10, 

£B  =  ±5,  ±4, 

4?/2  —  xy  =  —1. 

y  =  ±1,  ±|. 

35. 

a*2  +  xy  —  2y2  =  —44, 

a*  =  ±14,  ±1, 

*2/  +  3?/2  =      80. 

2/  =  =f   8,  ±5. 

36. 

a*2  +  3a*?/  =     54, 

a*  =  ±3,  ±36, 

xy  +  4?/2  =  115. 

y  =  ±5,  =FHf 

37. 

x(x  +  y)  =  40, 

as  =  ±5,  ±4*/2, 

y(aj  —  y)  =     6. 

2/  -  ±3,  ±   «/2. 

38. 

«(» + y) + y  (*  -  2/)  =  158, 

a*=±9,  ±8\/2, 

7a*(a*  +  ?/)  =    72?/(a* 

-?/).      y=±7,  ±   V2. 

39.  a*  +  2/  =  4,  a*4+?/4  =      82.  a>  =  3,  1 ;  y  =  1,  3. 

40.  x  -  y  =  3,  x5  -  yh  =  3093,  a*  =  5,  -2  ;  ?/  =  2,  -5. 

41.  a*2?/2  +  13a*?/  +12  =  0,  a*  =      4,  -3,  *  ±  ^D. 

X  +  y  =1.  y  =   -3,        4,  ^-^. 


EXAMPLES.  295 

5  ±  vT7 


2       ' 
5  =F  Vl7 


42.  a;  +  ?/  =    5,  ^Ltis.  a?  =      6,-1, 

4xy  =  12  -  a;2?/2.  y  =  —  1,      6, 

43.  x2y2  +  5a#  =  84,  a?  =  7,  1,  4  ±  ^28, 

x  4-  y  =     8.  y  =  1,  7,  4  T  ^28. 

44.  z2  +  4f  +  80  =  15a;  +  SOy,  x  =  4,  3,  6,  2, 

^  =6.  y=  },  2,  1,3. 

45.  9s2  +  ?/2-  63s-  21y+  128  =  0,       x  =  2,  |,  4,    -§, 

a?y=  4.       ?/  =  2,  6,  1,  12. 

46.  a4  +  2/4  =  14*y,  a  =  ^((1  ±  \/3),  ?/l  ±  -IV 

x+y=a.  ^-fa  TVS).  Kit  A). 

47.  Find  two  numbers  whose  difference  added  to  the 
difference  of  their  squares  is  14,  and  whose  sum  added  to 
the  sum  of  their  squares  is  26.  Ans.  4,  2. 

48.  Find  two  numbers  such  that  twice  the  first  with  three 
times  the  second  may  make  60,  and  twice  the  square  of  the 
first  with  three  times  the  square  of  the  second  may  make 
840.  Ans.  18  and  8,  or  6  and  16. 

49.  Find  two  numbers  whose  sum  is  nine  times  their 
difference,  and  whose  product  diminished  by  the  greater 
number  is  equal  to  twelve  times  the  greater  number  divided 
by  the  less.  Ans.  5,  4. 

50.  Find  two  numbers  whose  difference  multiplied  by  the 
difference  of  their  squares  is  32,  and  whose  sum  multiplied 
by  the  sum  of  their  squares  is  272.  Ans.  5,  3. 

51.  Find  two  numbers  whose  product  is  equal  to  their 
sum,  and  whose  sum  added  to  the  sum  of  their  squares  is  12. 

Ans.  2,  2. 

52.  Find  two  numbers  whose  sum  added  to  their  product 
is  34,  and  the  sum  of  whose  squares  diminished  by  their 
sum  is  42.  Ans.  4,  6. 


296  EXAMPLE  8. 

5?.  A  number  consisting  of  two  digits  has  one  decimal 
place ;  the  difference  of  the  squares  of  the  digits  is  20,  and 
if  the  digits  be  reversed,  the  sum  of  the  two  numbers  is  11  : 
find  the  number.  Ans.  6.4,  or  4.6. 

54.  A  man  has  to  travel  a  certain  distance  ;  and  when  he 
has  traveled  40  miles  he  increases  his  speed  2  miles  per 
hour.  If  he  had  traveled  with  his  increased  speed  during 
the  whole  journey  he  would  have  arrived  40  minutes  earlier ; 
but  if  he  had  continued  at  his  original  speed  he  would  have 
arrived  20  minutes  later.  Find  the  whole  distance  he  had  to 
travel,  and  his  original  speed.  Ans.  60,  10. 

55.  A  and  B  are  two  towns  situated  18  miles  apart  on  the 
same  bank  of  a  river.  A  man  goes  from  A  to  B  in  4  hours, 
by  rowing  the  first  half  of  the  distance  and  walking  the 
second  half.  In  returning  he  walks  the  first  half  at  the  same 
rate  as  before,  but  the  stream  being  with  him,  he  rows  1J 
miles  per  hour  more  than  in  going,  and  accomplishes  the 
whole  distance  in  3^  hours.  Find  his  rates  of  walking  and 
rowing.  Ans.  4J  walking,  4^  rowing  at  first. 

56.  A  and  B  run  a  race  round  a  two  mile  course.  In  the 
first  heat  B  reaches  the  winning  post  2  minutes  before  A. 
In  the  second  heat  A  increases  his  speed  2  miles  per  hour, 
and  B  diminishes  his  as  much ;  and  A  then  arrives  at  the 
winning  post  2  minutes  before  B.  Find  at  what  rate  each 
man  ran  in  the  first  heat.  Ans.  10,  12  miles  per  hour. 

57.  Find  two  numbers  whose  product  is  equal  to  the  dif- 
ference of  their  squares,  and  the  sum  of  their  squares  equal 
to  the  difference  of  their  cubes.  Ans.  £0),  1(5  +  0>). 

58.  The  fore-wheel  of  a  coach  makes  6  revolutions  more 
than  the  hind-wheel  in  going  120  yards ;  but  if  the  circum- 
ference of  each  wheel  be  increased  1  yard,  the  fore-wheel 
will  make  only  4  revolutions  more  than  the  hind-wheel  in  the 
same  distance  :  find  the  circumference  of  each  wheel. 

Ana  4  and  5  yards 


RATIO  —  DEFINITIONS.  207 


CHAPTER     XV. 

RATIO  — PROPORTION  — VARIATION. 

150.  Ratio  —  Definitions.  —  The  relative  magnitude  of 
two  quantities,  measured  by  the  number  of  times  which  the 
first  contains  the  second,  is  called  their  Ratio. 

The  ratio  of  a  to  b  is  usually  written  a :  b  ;  a  is  called 
the  first  term,  and  b  the  second  term  of  the  ratio.  The  first 
term  is  often  called  the  antecedent,  and  the  second  term  the 
consequent. 

Magnitudes  must  always  be  expressed  by  means  of  num- 
bers; and  the  number  of  times  which  one  number  contains 
the  other  is  found  by  dividing  the  one  by  the  other.     Hence 

the  ratio  a  :  b  may  be  measured  by  the  fraction  -. 

Thus,  the  ratio  a  :  b  is  equal  to  -,  or  is  -. 

b  b 

Concrete  quantities  of  different  kinds  can  have  no  ratio  to 
one  another ;  thus,  we  cannot  compare  pounds  with  yards, 
or  dollars  with  days. 

To  compare  two  quantities,  they  must  be  expressed  in 
terms  of  the  same  unit.  For  example,  the  ratio  of  4  yards 
to  15  inches  is  measured  by  the  fraction 

4  x  3  x  12        48 
15  5~* 

A  ratio  is  called  a  ratio  of  greater  inequality,  of  less  in- 
equality, or  of  equality,  according  as  the  antecedent  is  greater 
than,  less  than',  or  equal  to  the  consequent. 

Ratios  are  compounded  by  multiplying  together  the  ante- 
cedents of  the  given  ratios  for  a  new  antecedent,  and  the 


298  RA  TIO  —  DEFINITIONS. 

consequents  for  a  new  consequent.  Thus,  the  ratio  com- 
pounded of  the  three  ratios, 

3a  :  26,  4a6  :  5c2,  c  :  a, 

is  3a  X  Aab  x  c :  26  x  5c2  x  a,  or  6a :  5c0 

When  the  ratio  a :  6  is  compounded  with  itself,  the  result- 
ing ratio  is  a2 :  6'2,  and  is  called  the  duplicate  ratio  of  a :  6. 
Similarly,  the  ratio  a3 :  63  is  called  the  triplicate  ratio  of  a :  6. 
Also  the  ratio  a^ :  6-  is  called  the  subcluplicate  ratio  of  a:  bo 

If  we  interchange  the  terms  of  a  ratio,  the  result  is  called 
the  inverse  ratio.     Thus 

b  :  a  is  the  inverse  of  a  :  b. 

The  inverse  ratio  is  the  reciprocal  of  the  direct  ratio. 

When  the  ratio  of  two  quantities  can  be  exactly  expressed 
by  the  ratio  of  two  integers,  the  quantities  are  said  to  be 
commensurable;  when  the  ratio  cannot  be  exactly  expressed 
by  the  ratio  of  two  integers,  they  are  said  to  be  incommen- 
surable. 

Although  we  cannot  find  two  integers  which  will  exactly 
measure  the  ratio  of  two  incommensurable  quantities,  yet  we 
can  always  find  two  integers  whose  ratio  differs  from  the 
required  ratio  by  as  small  a  quantity  as  we  please. 

For  example,  the  ratio  of  a  diagonal  to  a  side  of  a  square 
cannot  be  exactly  expressed  by  the  ratio  of  two  whole 
numbers,  for  this  ratio  is  y/2,  and  we  cannot  find  any 
fraction  which  is  exactly  equal  to  y/2  ;  but  by  taking  a  suf- 
ficient number  of  decimals,  we  may  find  y/2  to  any  required 
degree  of  approximation.     Thus 

y/2  =  1.4142135 

and  therefore     y/2  >  gUlU  and  <  MI±M±. 

That  is,  the  ratio  of  a  diagonal  to  a  side  of  a  square  lies 

between  ioyJ-ooo  an<^  ioooooo*  ai1^  therefore  differs  from 
either  of  these  ratios  by  less  than  one-millionta ;  and  since 
the  decimals  may  be  continued  without  end  in  extracting  t\w 


PROPERTIES    OF   RATIOS.  293 

square  root  of  2,  it  is  evident  that  this  ratio  can  be  expressed 

as  a  fraction  with  an  error  less  than  any  assignable  quantity. 

In  general.    When  a  and  b  are  incommensurable,  divide  b 

into  n  equal  parts  each  equal  to  x,  so  that  b  =  nx.  where  n 

is  a   positive  integer.     Also  let  a  >  mx,  but  <  (m  -f-  l)x\ 

then 

a     mx       ,      (/ii4-  l)x 

->  —  and  <  - — — —  ; 

b      nx  nx 

that  is,  -  lies  between  —  and  —  ;  so  that  -  differs  from 

b  n  n  b 

971  1 

—  by  a  quantity  less  than  -.     And  since  we  can  choose  x 
n  n 

(our  unit  of  measure)  as  small  as  we  please,  n  can  be  made 

as  great  as  we  please,  and  therefore  -  can  be  made  as  small 

n 

as  we  please.     Hence  two  integers,  m  and  n,  can  be  found 

whose  ratio  will  express  the  ratio  a  :  b  to  any  required  degree 

of  accuracy. 

Note.  — The  student  should  observe  that  the  Algebraic  definition 
of  ratio  deals  with  numbers,  or  with  magnitudes  represented  by 
numbers,  while  the  Geometric  definition  of  ratio  deals  with  concrete 
magnitudes,  such  as  lines  or  areas  represented  Geometrically,  but  not 
referred  to  any  common  unit  of  measure. 

151.  Properties  of  Ratios. —  (1)  If  the  terms  of  a 
ratio  be  multiplied  or  divided  by  the  same  number  the  value  of 
the  ratio  is  unaltered. 

Por  -  =  —  (Art.  /9). 

b       mb  K  J 

Thus  the  ratios  2:3,  6:9,  and  2m :  3m,  are  all  equal  to 
each  other. 

Two  or  more  ratios  are  compared  by  reducing  the  fractions 

which  measure  them  to  a  common  denominator.     Thus,  sup- 

,        t        7  rr^i       ci        ad     c        be , 

pose  a  :  b  and  c  :  d  are  two  ratios,      ihen  -  =  -— ,    -  =  —  ; 

b         bd     d        bd 

hence  the  ratio  a:b  >,  = ,  or  <  the  ratio  c :  d,  according  as 

ad  >,  = ,  or  <  be. 


300  PROPERTIES    OF  RATIOS. 

The  ratio  of  two  fractions  can  be  expressed  as  a  ratio  of 


Thus  the  ratio  -  :  -  is  measured  by  the   fraction  -  or  —  ; 
b  d  c         be 

and  is  therefore  equivalent  to  the  ratio  ad :  be. 

(2)  A  ratio  of  greater  inequality  is  diminished,  and  a  ratio 
of  less  inequality  is  increased,  by  adding  the  same  quantity  to 
each  of  its  terms;  that  is,  the  ratio  is  made  more  nearly  equal 
to  unity. 

Let  a :  b  be  the  ratio,  and  let  a  +  x :  b  +  x  be  the  new 
ratio  formed  by  adding  x  to  each  of  its  terms. 

Then  -  -  a  +  X  =  ^  ~  ^  ; 

b       b  -h  x       b(b  -h  x}1 

and  a  —  b  is  positive  or  negative  according  as  a  is  greater 

or  less  than  b. 

Hence  a  +  x  <,  or  >  -    according,  as  a  >,  or  <  b  ;  that  is, 
b  -f  x  b 

the  resulting  ratio  is  brought  nearer  to  unity. 

For  example,  if  to  each  term  of  the  ratio  3  :  2  we  add  12, 
the  new  ratio  15  :  14  is  less  than  the  former,  because  Jf  =  ly1^ 
is  clearly  less  than  §  =  1|. 

Also,  if  to  each  term  of  the  ratio  2  :  3  we  add  12,  the  new 
ratio  14:  15  is  greater  than  the  former,  since  y|  is  clearly 
greater  than  }. 

(3)  Similarly,  it  can  be  proved  that  a  ratio  of  greater  in- 
equality is  increased,  and  a  ratio  of  less  inequality  is  dimin- 
ished, by  taking  the  same  quantity  from  both  its  terms. 

(4)  The  following  is  a  very  important  proposition  concern- 
ing equal  ratios. 

If^  =  -  =  -= ,  then  each  of  these  ratios 

b       d       f 


_  lpan  +  qcn  +  ren  +  .  .  .  .V 

~  \2)bn  +  qd»  +  rf*  + / 

where  p,  q,  r,  n  are  any  quantities  whatever* 


EXAMPLES.  301 

het*=c=«   = =  k. 

b       d       f 

then     a  =  bk,  c  =  dk,  e  =  fk, ; 

therefore  pan+qcn  +  ren  +  ....  =pbnkn+qdnkn  +  rfnkn+  .  .  „  . 

\P&+  q&+  rfn  + )  ~  b      d     f      {  } 

By  giving  different  values  to  p,  g,  r,  w,  many  particular 
cases  of  this  general  proposition  may  be  deduced  ;  or  they 
may  be  proved  independently  by  the  above  method. 
Suppose  b=1,  then  we  have  from  (1) 

a  _  c  _  e  _                  _  pa  +  qc  +  re  -f-  .  .  .  .        ,^\ 
b~  d~  f~ -~fb  +  qd  +  ff+ '   {  } 

Suppose  n  =  1,  and  p  =  q  =  r  = ,  then  (1)  becomes 

a  _  c  _  e  _                _  r t  +  c  +  e  .  .  .  .  ,ox 

b~d~~f~  &  +  <*+/ '   K) 

That  is,  when  a  series  of  fractions  are  equal,  each  of  them  is 
equal  to  the  sum  of  all  the  numerators  divided  by  the  sum  of 
cdl  the  denominators. 

EXAMPLES. 

1.  If  -  =  |,  find  the  value  of  — -. 

y  7x  -{-  2y 

DX  q 

5x  -  3y        y __  y  -  3  __    3 

7x  +  2y       7x  +  2        2J  +  2 
2/ 

2.  If  a :  5  be  in  the  duplicate  ratio  of  a  +  re :  6  -f-  x,  prove 
that  jc2  =  ah. 


From  the  condition  , 


la  +  a?\2  _  a 
V&  +  a/    ="  &' 

.     a2b  +  2a6a?  +  bx2  =  ad2  -f-  2a6aj  +  aa?2. 

.-.     x2  =  ab. 


302  PROPORTION  —  DEFINITIONS. 

Find  the  ratio  compounded  of 

3.  The  ratios  4  :  15  and  25  :  36.  Ans.  5  :  27. 

4.  The  ratio  27  :  8,  and  the  duplicate  ratio  of  4  :  3.      6:1. 

5.  The  ratio  169  :  200,  and  the  duplicate  ratio  of  15  :  26. 

Ans.  9  :  32. 

6.  If  4#2  +  y1  —  4xy,  find  the  ratio  x :  y.  1:2. 

7.  What  is  the  ratio  x :  y,  if  the  ratio  Ax  +  by :  3x  —  y  is 
equal  to  2  ?  Ans.  7  :  2. 

8.  If  Ix  —  4y :  3x  +  y  =  5  :  13,  find  the  ratio  x :  y. 

Ans.  3:4. 

PROPORTION. 

152.    Definitions. — Four  quantities  are  said    to   be   in 

proportion  when  the  ratio  of  the  first  to  the  second  is  equal 

to  the  ratio  of  the  third  to  the  fourth ;  and  the  terms  of  the 

ratios  are  said  to  be  proportionals. 

a       c 
Thus,  if  -  =  -,  then  a,  &,  c,  c7,  are  called  proportionals, 

or  are  said  to  be  in  proportion.     The  proportion  is  written 

a:  b  =  c  :  d, 
or  a :  b  ::  c  :  d, 

which  is  read  "  a  is  to  b  as  c  is  to  d." 

The  Algebraic  test  of  a  proportion  is  that  the  two  fractions 
which  represent  the  ratios  shall  be  equal. 

The  four  terms  of  the  two  equal  ratios  are  called  the  terms 
of  the  proportion.  The  first  and  fourth  terms  are  called  the 
extremes,  and  the  second  and  third,  the  means.  Thus,  in 
the  above  proportion,  a  and  d  are  the  extremes  and  b  and  c 
the  means. 

Quantities  are  said  to  be  in  continued  proportion  when  the 
first  is  to  the  second,  as  the  second  is  to  the  third,  as  the 
third  to  the  fourth,  and  so  on.  Thus  a,  b,  c,  (/,  e,  /,  .  .  . 
are  in  continued  proportion  when 

a:b  =  b:c  =  c:  d  =  d;  e  =  e:f  = 


PROPERTIES    OF  PROPORTIONS.  303 

If  a,  b,  c,  be  in  continued  proportion,  b  is  said  to  be  a 
mean  proportional  between  a  and  c ;  and  c  is  said  to  be 
a  third  proportional  to  a  and  b. 

If  «,  b,  c,  d  be  in  continued  proportion,  b  and  c  are  said 
to  be  two  mean  proportionals  between  a  and  d ;  and  so  on. 

153.  Properties  of  Proportions.  —  ( 1 )  If  four  quan- 
tities are  in  proportion,  the  product  of  the  extremes  is  equal 
to  the  product  of  the  means. 

Let  the  proportion  be  a  :  b  =  c  :  d. 

Then  by  definition  (Art.  156),  -  =  -. 

b       d 

Multiplying  by  bd,  ad  =  be (1) 

Hence  if  any  three  terms  of  a  proportion  are  given,  the 

fourth  may  be  found  from  the  relation  ad  =  be. 

Note.  —  This  proposition  furnishes  a  more  convenient  test  of  a 
proportion  than  the  one  in  Art.  152.  Thus,  to  ascertain  whether 
2  :  5  : :  6  :  16,  it  is  only  necessary  to  compare  the  product  of  the  means 
and  extremes;  and  since  5  x  6  is  not  equal  to  2  x  16,  we  see  that 
2,  5,  6,  16,  are  not  in  proportion. 

If  b  =  c,  we  have  from  (1),  ad  =  b2 ;     .*.  b  =  sfac. 

That  is,  the  mean  proportional  between  tivo  given  quanti- 
ties is  equal  to  the  square  root  of  their  product. 

(2)  Conversely,  If  the  product  of  two  quantities  be  equal 
to  the  product  of  tivo  others,  tivo  of  them  may  be  made  the 
extremes,  and  the  other  two  the  means,  of  a  proportion. 

For  let  ad  =  be. 

Dividing  by  bd,  ^  =  -  ; 

that  is,  a  :  b  :  :  c  :  cl. 

In  a  similar  manner  it  "may  be  shown  that  the  proportions 

a  :  c  :  :  b  :  d, 

b:  a  :  :  d  :  c, 

bid  : :  a:  c, 

c  :  d  : :  a  :  b,  etc., 
are  all  true  provided  that  ad  =  be. 


304  PROPERTIES   OF  PROPORTIONS. 

If  four  quantities  are  in  proportion  they  will  be  in  propor- 
tion by 

(3)  Inversion.  —  If  a  :  b  : :  c :  d,  then  b  :  a  : :  d :  c. 

For  -  =  -  ;  therefore  1  --  -  =  1  --  - ; 
b       d  b  d 

that  is,  -  =  -  ;  or  b  :  a  : :  die. 
a       c 

(4)  Alternation.  —  If  a  :  b  : :  c:  d,  then  a:  c  : :  b :  d. 

For  ad  =  be  ;  therefore  -—  =  —-; 
ca       ca 

that  is  -  =  -  ;  or  a  :  c  : :  b  :  d. 
c        d 

(5)  Composition.  —  If  a :  b : :  c :  d,  then  a-{-b:b::  c+d :  d. 

For  -  =  - ;  therefore  -  +  1  =  -  +  1  ; 
b       d  b  d 

that  is  — ! —  =  — ! ;  or  a  +  o  :  b  : :  c  -f  d  :  a. 

b  d 

(6)  Division.  —  If  a  :  b  :  :  c  :  cZ,  then  a  —  b  :  b  :  :  c  —  d  :  d. 

For  -  =  -  ;  therefore 1  = 1  ; 

&       d  6  d 

.u  .  .   a  —  6       c  —  d  ,,  ,, 

that  is =  ;  or  a  —  6  :  b  : :  c  —  d :  d. 

b  d 

In  a  similar  manner  it  may  be  shown  that  the  sum  (or  the 

difference)  of  the  first  and  second  of  two  quantities  is  to  the 

first  as  the  sum  (or  the  difference)  of  the  third  and  fourth 

is  to  the  third. 

(7)  Composition  and  Division.  — If  a  :  b  ::  c:  d,  then 

a  -\-  b:  a  —  b  ::  c  +  d :  c  —  d. 
For  by  (5)  and  (G), 

a  +  b       c  +  d         a  —  b       c  —  d 


b                d     '             b 

cZ 

division, 

a  +  b  _  c  -f-  d . 

a  —  b       c  —  d 

fj,  +  b  :  a  —  b  :  :  c  +  cj :  c 

-  e*. 

PROPERTIES    OF  PROPORTIONS.  305 

(8)  If  three  quantities  are  in  continued  proportion ,  the  first 
is  to  the  third  in  the  duplicate  ratio  of  the  first  to  the  second. 

For  if  a  :  b  :  :  b  :  c,  then  -  =  -. 
b        c 

^       a       a       b       a       a       a2 

c       b      c      b      b      b2 

Hence  a:  c  :  :  a2 :  b2. 

Similarly  it  may  be  shown  that  if  a  :  b  : :  b  :  c  : :  c:d, 
then  a  :  d  :  :  a3 :  b3. 

(9)  Quantities  ivhich  are  proportional  to  the  same  quanti- 
ties, are  proportional  to  each  other. 

If  a:b::e:f,  and  c:  d:  :  e  :  f,  then  ty:  b : :  c :  d. 

v      a        e         ,  c        e      .,        -        a        c 
ror  -  =  — ,  and  -  =  — :  therefore  -  =  -, 
&      /         d      /'  &      d' 

or  a  :  6  :  :  c  :  d. 

(10)  77*  e  products  of  the  corresponding  terms  of  two  or 
more  proportions  are  in  proportion. 

For  if  a  :  & : :  c :  (7,  and  e  :  /  :  :  g  :  h, 

then  —  =  — ,     and     —  =  ": 

b       d  f       h 

therefore  —  =  —  ;     or     ae :  bf : :  eg :  dh. 

bf       dh 

(11)  When  four  quantities  are  in  proportion,  if  the  first 
and  second  be  multiplied,  or  divided,  by  any  quantity,  as 
also  the  third  and  fourth,  the  resulting  quantities  will  be  in 
proportion. 

For  if  a :  6 : :  c :  d,  then  -  =  - ; 
'     b       cf 

therefore        —  =  —  ;     or     ma  :  wo  : :  nc  :  w«. 
?»&        nd 

Similarly  it  may  be  shown  that  —:  —  ::-:-. 

m     m       n    n 


306  PROPERTIES   OF  PROPORTIONS. 

(12)  In  a  similar  manner  it  may  be  shown  that  if  the  first 
and  third  terms  be  multiplied,  or  divided,  by  any  quantity, 
and  also  the  second  and  fourth,  the  resulting  quantities  will 
be  in  proportion. 

(13)  If  four  quantities  are  in  proportion,  the  like  powers, 

or  roots,  of  these  quantities  will  be  in  proportion. 

a       c 
For  if  a  :  b  : :  c  :  d,  then  -  =  - ; 
b       a 

therefore         ?-  =  ^ ;         .'.     an  :  bn  :  :  cn :  d\ 


an  _  cn . 
b"  ~  dn ' 

•'• 

an:br 

i         i 
an  _  cn  m 

. 

an:  b 

Also  LJ7=n-;         •'•     a«:b~n::c»:dk 

b~n       dn 

(14)  If  any  number  of  quantities  are  in  proportion,  any 
antecedent  is  to  its  consequent,  as  the  sum  of  all  the  antecedents 
is  to  the  sum  of  all  the  consequents. 

For  if  a  :  b  :  :  c  :  d  : :  e  :  f, 
then  by  (1),  ad  =  be,  and  of  =  be ;  also  «6  =  ba. 
Adding  a(6  +  cZ  +  /)  =  b(a  +  c  +  e); 

therefore  by  (2) ,  a  :  6  : :  a  +  c  +  e  :  6  -f  d  +  /. 

This  also  follows  directly  from  (3)  of  Art.  151. 

(15)  When  -,-,—, are  unequal,  it  follows  from 

b   d   f 

Ex.  3  of  Art.  106,  that 

a  +  c  +  e  +  g-}-..  .  .  .  . 


b  +  d  +  f  +  h+ 

is  greater  than  the  least,  and  less  than  the  greatest,  of  the 

„      ..        a    c    e    q 
fractions 


6'  d'  /  h' 

It  is  obvious  from  the  preceding  propositions  that  if  four 
quantities  are  in  proportion,  many  other  proportions  may  be 
derived  from  them.  The  propositions  jnst  proved  are  often 
useful  in  solving  problems.  In  particular,  the  solution  of 
certain  equations  is  greatly  facilitated  by  a  skilful  use  of  the 
operations  of  composition  and  division. 


EXAMPLES.  307 

EXAMPLES. 
1.    If  a  :  b  ::  e  :  d, 

show  that  a2  +  ab  :  c2  +  cd  : :  62  —  2a6  :  a*2  —  2cd. 

a       c 
Let  -  =  -  =  x ;  then  a  =  bx,  and  c  =  dx. 
b       d 


a2  +  ab        b2x2  +  b2x        b2 

c2  +  cd        d-x2  +  (J*u       d2' 

Also 

52  _  2a6        b2  -  2b2x        b2 
d2  -  2cd       d2  -  2d2x       d2' 

Therefore  by  (9),  a2  +  ab  :  c2  +  cd  : :  b2  -  2ab  :  d2  -  2cd. 

2     If  3a  +  66  +  c  +  2d  __  3a  +  6b  —  c  —  2d 

3a  - 

-  66  +  c  -  2d       3a  -  66  -  c  +  2d 

prove  that 

a  :  6  :  :  c  :  d. 

By  (7) 

2(3a  +  c)          2(3a  -  c) 
2(66  +  2d)        2(66  -  2d) 

By  (4) 

3a  +  c       66  -f  2d 
3a  —  c       66  -  2d 

Again  by  (7) 

,                      —  =  —           .*.     a  :  0  : :  c  :  a. 

'                      2c        4d 

3.    Find  a 

fourth  proportional  to  a3,  ary,  bx2y.       Ans.  by2. 

4.  Find  a  mean  proportional  between  1203s  and  3a3. 

An&.  Gcrx. 

5.  Find  a  third  proportional  to  Xs  and  2x2.  4x. 

6.  If  a  :  6  :  :  c  :  d,  show  that 

(i)    OC  :   6d  ::  c2 :  d2. 

(2)  oft  :  cd   ::  a2:  c2. 

(3)  a2   :  c2    :  :  a2  -  62  :  c2  -  d2. 
7.    If  a  :  6  :  :  c  :  d,  prove  that 

(1)  ab  +  cd  :  ab  -cd::  a2  +  c2 :  a2  -  c2. 

(2)  a2  +  ac  +  c2  :  a2  -  ac  +  c2  : :  62  +  bd  +  d2 :  62  -  bd  +  d2. 

(3)  a  :  6  :  :  V^a2  +  5c2 :  Vfed*  +  5d2. 

(4)  a  +  6 :  c  +  d : :  vV2  +  62 :  Vo2  +  ci2. 


308  VARIA  TION  —  DEFINITION. 

8.  If  'a  :  b  : :  c  :  d  :  j  e  -:  /,  prove  that 

2a2  +  3c2  -  5e2  :  262  +  3d2  -  5/2  : :  ae  :  6/. 

n     c  .       .,  ,.      a;2  +  a;  —  2       4.t2  +  5s  —  6 

9.  Solve  the  equation '- =  . 

1  x  —  2  bx  —  6 

^b*s.  x  =  0,  —2. 

10.  Find  x  in  terms  of  ?/  from  the  proportions  x :  y : :  a3 :  63, 
and  a  :  6  : :  Vc  +  a>  :  \/d  +  2/.  ^,IS.  )T  —  £« 

11.  If  a,  6,  c,  cZ  are  in  continued  proportion,  prove  that 

aid::  a3  +  bs  +  c3  :  63  +  c3  +  d8. 

VARIATION. 

154.  Definition.  —  One  quantity  is  said  to  vary  directly 
as  another  when  the  two  quantities  depend  upon  each  other 
in  such  a  manner  that  if  one  be  changed  the  other  is  changed 
in  the  same  proportion. 

Thus,  if  a  train  moving  uniformly,  travels  40  miles  in  an 
hour,  it  will  travel  80  miles  in  2  hours,  120  miles  in  3  hours, 
and  so  on  ;  the  distance  in  each  case  being  increased  or 
diminished  in  the  same  ratio  as  the  time.  This  is  expressed 
by  saying  that  when  the  velocity  is  uniform,  the  distance  is 
jwojiortional  to  the  time,  or  more  briefly,  the  distance  varies 
as  the  time.  We  may  express  this  result  with  Algebraic 
symbols  thus :  let  A  and  a  be  the  numbers  which  represent 
the  distances  traveled  by  the  train  in  the  times  represented 
by  the  numbers  B  and  b ;  that  is,  when  A  is  changed  to  any 
other  value  a,  B  must  be  changed  to  another  value  6,  so  that 
A  :  a  : :  B  :  b  ;  then  A  is  said  to  vary  directly  as  J5,  or 
more  briefly,  to  vary  as  B. 

Another  phrase,*  which  is  also  in  use,  is  "A  is  proportional 
to  Br 

*  Strictly  Bpeaking,  this  phrase  is  better  than  the  one  "  varies  as,"  which  is 
somewhat  antiquated  ;  but  in  deference  to  usage  we  retain  it.  The  student  must  not 
suppose  that  the  variation  here  considered  is  the  only  kind.  We  are  not  here 
concerned  with  variation  in  general,  but  merely  with  the  simplest  of  all  the  possible 
kinds  of  variation. 


DIFFERENT   CASES   OF    VARIATION.  30i) 

This  relation  is  sometimes  expressed  by  the  symbol  oc,  so 
that  A  oc  B  is  read  c '  A  varies  as  B. ' ' 

It  will  thus  be  seen  that  variation  is  merely  an  abridged 
method  of  expressing  proportion,  and  that  four  quantities 
are  understood  though  only  two  are  expressed. 

If  A  varies  as  Z?,  then  A  is  equal  to  B  multiplied  by  some 
constant  quantity. 

For  suppose  that  a,  av  a2,  .  .  .  .  ,  b,  bv  62,  .  .  =  .  .  are 
corresponding  values  of  A  and  B. 

Let  a  and  b  denote  one  pair  of  these  values,  so  that  when 
A  has  the  value  a,  B  has  the  value  b  ;  then  we  have  by  the 
definition,  A  :  a  : :  B  :  6.     Hence 

A  =  -B  =  mB, 
b 

where  m  is  equal  to  the  constant  ratio  a  :  b. 

155.  Different  Cases  of  Variation.  —  TJiere  are  four 
different  kinds  of  variation. 

(1)  One  quantity  is  said  to  vary  Directly  as  another  when 
the  two  increase  or  decrease  together  in  the  same  ratio. 
Thus, 

A  cc  B,  or  A  =  mB  (Art.  154). 

For  example,  If  a  man  works  for  a  certain  sum  per  hour, 
the  amount  of  his  wages  varies  as  the  number  of  hours 
during  which  he  works. 

(2)  One  quantity  is  said  to  vary  Inversely  as  another  when 
the  first  varies  as  the  reciprocal  of  the  other.  Thus  A  varies 
inversely  as  B  is  written 

Aoo  —,  or  A  =  —,  where  m  is  a  constant. 
B  B 

For  example,  If  a  man  has  to  perform  a  certain  journey, 
the  time  in  which  he  will  perform  it  varies  inversely  as  his 
speed.  If  he  doubles  his  speed,  he  will  go  in  half  the  time  ; 
and  so  on. 


310  PROPOSITIONS   IN    VARIATION. 

(3)  One  quantity  is  said  to  vary  as  two  others  Jointly, 
when  the  first  varies  as  the  product  of  the  other  two.  Thus 
A  varies  as  B  and  G  jointly  is  written 

A  oc  BC,  or  A  =  mBG,  where  m  is  a  constant. 

For  example,  The  wages  to  be  received  by  a  workman  will 
vary  as  the  number  of  days  he  has  worked  and  the  wages  per 
day  jointly. 

(4)  One  quantity  is  said  to  vary  Directly  as  a  second  and 
Inversely  as  a  third,  when  it  varies  jointly  as  the  second  and 
the  reciprocal  of  the  third.  Thus  A  varies  directly  as  B 
and  inversely  as  C  is  written 

7?  ~R 

A  oc  — ,  or  A  =  ra  — ,  where  m  is  a  constant. 
G  G 

For  example,  The  base  of  a  triangle  varies  directly  as  the 
area  and  inversely  as  the  altitude. 

In  the  different  cases  of  variation  just  defined,  to  deter- 
mine the  constant  m  it  will  only  be  necessary  to  have  given 
one  set  of  corresponding  values. 

Example  1.    If  A  oc  B,  and  A  =  3  when  B  =  12,  we  have 

A  =  mB\         .-.     3    =  m  x  12  ; 

or  ra  =  J  ;  .-.     A  =  \B. 

2.  If  A  varies  as  B  and  inversely  as  (7,  and  A  =  6  when 
B  =  2  and  C  =  9,  we  have 

A  =  m-  ;         .-.     G   =  m  x  f  ; 
C 

or  m  =  27;  .-.     A  =  27^. 

G 

156.  Propositions  in  Variation.  —  The  simplest  method 
of  treating  valuations  is  to  convert  them  into  equations. 

(1)  If  A  oc  J5,  and  75  oc  C,  then  iaC. 

For  let      A  =  ??i75,  and  5  =  nC  (Art.  154), 
where  ra  and  n  are  constants. 


PROPOSITIONS   IN    VARIATION.  31 1 

Then  A  =  mnC ; 

.*.     A  oc  (7,  since  ran  is  constant. 

In  like  manner,  if  A  cc  B,  and  Z?  cc  — ,  then  ^4  cc  — . 

c  c 

(2)   If  .1  oc  C,  and  5aC,  then  .1  ±  B  cc  C,  and  \/Ze  cc  C. 

For  let  A  =  mC,  and  Z>  =  ?iC, 

where  m  and  ?i  are  constants. 
Then  A  ±  B  =  (m  ±  n)C. 

.*.     .4  ±  B  cc  C\  since  m  ±  ?*  is  constant. 


Also  V^4£  =  Vm/iC"2  =  CVwmi. 

V-4I2  cc  C,  since  V"*/*  is  constant. 

(3)  If  A  cc  £C,  then  J5  cc  4  and  C  cc  -. 

For  let  ^1  =  mBC;  then  JB  =  i  -. 

m  (7 

.-.     5  cc  4.       Similarly  C  ex  A 
(J  B 

(4)  If  A*  B,  and  C  «  Z>,  then  AC  <x  3Z). 
For  let  ^4  =  m5,  and  C  =  ?iZ). 

Then  .4C  =  mnBD.        ,-.     ^10  oc  SZ>. 

(5)  If  A  cc  5,  then  .4*  cc  £!. 

For  let  A  =  ra£ ;  then  An  =  m"^. 

.-.      AnazB\ 

(6)  If  A  cc  B  ichen  C  is  constant,  and  Ace  C  ichen  B  is 
constant,  then  A  cc  BC  ichen  both  B  and  C  are  variable. 

The  variation  of  A  depends  on  the  variations  of  the  two 
quantities  B  and  C.  Suppose  these  latter  variations  to  take 
place  successively,  each  in  its  turn  producing  its  own  effect 
on  A. 

Let  then  B  be  changed  to  b,  and  in  consequence  let  A  be 
changed  to  a',  C  being  constant ;  then,  by  supposition, 

A=B 
a!       h' 


312  PROPOSITIONS  IN    VARIATION. 

Now  let  C  be  changed  to  c,  and  in  consequence  let  a'  be 
changed  to  «,  b  being  constant ;  then,  by  supposition, 
rf_  G 
a        c 

Hence  4  X  <*-  =  |  X  2, 

a       a        o        c 

or  d  =  BC.         ...     ^cciSC. 

a        oc 

The  following  are  illustrations  of  this  proposition. 

The  amount  of  work  done  by  a  given  number  of  men  varies 
directly  as  the  number  of  days  they  work,  and  the  amount 
of  work  done  in  a  given  time  varies  directly  as  the  number  of 
men ;  therefore  when  the  number  of  days  and  the  number 
of  men  are  both  variable,  the  amount  of  work  will  vary  as 
the  product  of  the  number  of  men  and  the  number  of  days. 

Again,  the  area  of  a  triangle  varies  directly  as  the  base 
when  the  height  is  constant,  and  directly  as  the  height  when 
the  base  is  constant ;  hence  when  both  the  base  and  the 
height  are  variable,  the  area  will  vary  as  the  product  of 
the  base  and  height. 

In  the  same  manner,  if  A  varies  as  each  of  any  number  of 
quantities,  B,  C,  D,  .  .  .  when  the  rest  are  constant,  then 
when  they  all  vary  A  varies  as  their  product.  Also,  the 
variations  may  be  either  direct  or  inverse. 

Note.  —  This  principle  is  interesting  because  of  its  frequent 
occurrence  in  Physical  Science.  For  example,  in  the  theory  of  gases 
it  is  found  by  experiment  that  the  pressure  p  of  a  gas  varies  as  the 
"absolute  temperature"  t  when  the  volume  v  is  constant,  and  that 
the  pressure  varies  inversely  as  the  volume  when  the  temperature  is 
constant;  that  is, 

p  cc  t,  when  v  is  constant; 

and  p  cc  -,  when  t  is  constant. 

v 

From  these  results,  we  should  expect  that,  when  both  t  and  v  vary, 
we  should  have  the  formula 

p  cc  -,  or  ^  =  a  constant, 
v         t 

which  by  actual  experiment  is  found  to  be  the  case. 


y  =  iy^f  =  go. 


PROPOSITIONS  IN    VARIATION.  313 

/l. /If  y  varies  inversely  as  #2  —  1,  and  is  equal  to  24  when 
cc  =  10,  find  y  when  x  =  5. 

Since  ?/ oc  — ,  y  =  — ,  by  (2)  of  Art.  159. 

X"  —  1  x    —  1 

As  y  =  24  when  x  =  10,  we  have 

24  =  — .         .-.     m  =  24  x  99. 

99 

Hence,  when  a:  =  5,  we  have 

24  x  99 
x*  -  1 

2.  The  pressure  of  a  gas  varies  jointly  as  its  density  and 
its  absolute  temperature  ;  also  when  the  densit\T  is  1  and  the 
temperature  300,  the  pressure  is  15.  Find  the  pressure 
when  the  density  is  3  and  the  temperature  is  320. 

Let  p  =  the  pressure,  t  =  the  temperature,  and  d  =  the 
density. 

Then,  since  p  oc  ft?,  we  have  p  =  mtel,  by  (3)  of  Art.  159. 
As  p  =  15  when  t  =  300  and  d  =  1,  we  have 
15  =  m  x  300  x  1.         .".     m  =  ^j. 
Hence,  when  d  =  3  and  £  =  320,  we  have 
p  =  ^  x  320  x  3  =  48. 

3.  The  time  of  a  railway  journey  varies  directly  as  the 
distance  and  inversely  as  the  velocity ;  the  velocity  varies 
directly  as  the  square  root  of  the  quantity  of  coal  used  per 
mile,  and  inversely  as  the  number  of  cars  in  the  train.  In  a 
journey  of  25  miles  in  half  an  hour  with  18  cars,  10  cwt.  of 
coal  is  required :  how  much  coal  will  be  consumed  in  a 
journey  of  21  miles  in  28  minutes  with  16  cars? 

Let  t  =  the  time  in  hours,  d  =  the  distance  in  miles, 
v  =  the  velocity  in  miles  per  hour,  q  =  the  quantity  of  coal 
in  cwt.,  and  n  =  the  number  of  cars. 

Then  we  have  t  oc  -,  and  v  cc  — ^. 


v 


dn  .  dn 

_,     or     t  —  m—. 


314  EXAMPLES. 

As  d  =  25  when  t  =  f ,  n  =  18,  and  g  =  10,  we  have 

.  25  x  18  VlO  ,.^10     dn 

*  =  ra — — .       .'.    ra  = — ,    and   t=  — -. 

2  ^10  25  X  36'  25  X  36y/g 

Hence,  when  d  =  21,  2  =  f§,  and  n  =  16,  we  have 

28  =  v^io  x  21  x  16  =  y/Io  X  28 

25  X  36y/g  25  x  3^q' 

...    V5  =  ffi  *  28  *  60  =  WJQ. 
*  25  X  3  X  28  5 

.-.     g  =  ¥_  =  6f. 

Hence  the  quantity  of  coal  is  6|  cwt. 

4.  ^4  varies  as  .B,  and  A  is  5  when  B  is  3  ;   what  is  .4 
when  J3  is  5  ?  ^l?is.  8 J. 

5.  ^4  varies  inversely  as  5,  and  ^4  is  4  when  5  is  15  ; 
what  is  A  when  5  is  12?  Ans.  5. 

6.  If  x  oc  y  and  y  cc  z,  show  that  #2  cc  ?/2. 

7.  If  x  oc  -,  and  2/  oc  -,  prove  that  2  oc  n\ 

8.  If  #  oc  z  and  ?/  cc  z,  prove  that  x2  —  y2  <x  z\ 


EXAMPLES. 

Find  the  ratio  compounded  of 

1.  The  ratio  32  :  27,  and  the  triplicate  ratio  of  3:4. 

Ans.  1  :  2. 

2.  The  ratio  6  :  25,  and  the  snbduplicate  ratio  of  25  :  36. 

Ans.  1  :  5. 

3.  The  triplicate  ratio  ol  xiy,  and  the  ratio  2y2 1  3x2. 

Ans.  2x :  By. 

4.  If  x:  y  =  3|,  find  the  value  of  (x  -  By):  (2x  -  5y). 

Ans.  1  :  5. 

5.  If  -  =  },  and  -  =  #,  find  the  value  of    Sax  ~  6-?/ . 

b       4  V  Aby  —  lax 

Ans.  17:7. 

6.  Find  x  :  y,  having  given  x2  +  6y2  =  bxy.  2  or  3. 


EXAMPLES.  315 

7.  Find  two  numbers  in  the  ratio  of  5  to  6,  and  whose 
sum  is  121.  Ans.  55  and  G6. 

8.  For  what  value  of  x  will  the  ratio  15  +  x  :  17  +  x 
be  J?  Ans.  -13. 

9.  Find  x  in  order  that  x  +  1  :  £  -f  4  may  be  the  duplicate 
ratio  of  3  :  5.  -4n*.  11  :  16. 

10.  Two  numbers  are  in  the  ratio  of  4  :  5,  and  if  6  be 
taken  from  each,  the  ratio  is  that  of  3:4:  find  the  numbers. 

Ans.  24,  30. 

11.  Find  two  numbers  in  the  ratio  of  5  :  6,  such  that  their 
sum  has  to  the  difference  of  their  squares  the  ratio  of  1  :  7. 

Ans.  35  :  42. 

12.  Find  x  so  that  x  :  1   may  be  the  duplicate  of  the 
ratio  8  :  x.  Ans.  4. 

13.  If  2x  :  3#  be  in  the  duplicate  ratio  of  2x  —  m  :  3y  —  to, 
prove  that  to2  =  6xy. 

14.  II  A:  B  be  the  subduplicate  ratio  of  A  —  x  :  B  —  x, 
prove  that  x  =  AB  :  (A  4-  B) . 

lo.    Prove  tuat  if  -1 — ! A-  =  --— ] c  =  -3— ! — L,  each 

of   these   ratios   is   equal    to    1   +  x   :    1    -f-   ?/,    supposing 

ai    +    a2    +    a3  D0^    *°    ^e    ZCr0' 

.«     t»    a  -  6  6  —  c  c  —  a  a  +  6  +  c 

lb.     II    =    ■ =  =    : * 

aty  +  bx       bz  +  ca;       c^/  +  az        ax  +  &?/  4-  cz 
prove  that  each  of  these  ratios  =  1  :  x  +  y  +  2,  supposing 
a  4-  6  4-  c  not  to  be  zero. 

17.  Find  a  mean  proportional  between  a36  and  ab3. 

Ans.  a2b2. 

18.  Find  a  third  proportional  to  (a  —  b)2  and  oa  —  b2. 

Ans.  (a  +  &)2. 

19.  If  a:b::c:d,  prove  that 


(1)  2a  +  3c  :3a  4-  2c: 

(2)  /a  4-  to& :  pa  4-  76 : 

(3)  \/a^T^  :  V'-2  4-  d2 : 

(4)  a2c  4-  «c2 :  b2d  4-  &# : 


2b  4-  3c7 :  36  4-  2d. 
Ic  4-  ww2 :  pc  4-  ?<*• 


V«3  +  ^  :  V?  4-  d\ 
(a  4-  c)3:  (6  4-  d)\ 


316  EXAMPLES. 

Find  the  value  of  x  in  each  of  the  proportions : 

20.  3»  -  1  :  6x  -  7  : :  7x  -  10  :  9x  +  10.      ^4??s.  8  or  |. 

21.  x2  —  2x  -f-  3  :  2x  —  3  : :  x2  —  3x  +  5  :  3x  —  5.       2  or  0. 

22.  2z  —  1  :  x  +  4  : :  x2  -f  2a;  —  1  :  x2  +  a;  +  4.       5  or  0< 


23.  (V^+l+V^-l)  :  (VaM-I-V^-l)::4a;-l:2.        1J. 

24.  If  a :  b : :  c :  d : :  e  r/,  prove  that 

a3    _|_    c3    _|_    e3  .  &3    +    d3    +    y3  .  .  ace  .  ^ 

25.  If  a  :  b  :  :  c  :  d,  prove  that 

(1)  «(c  +  c?)  =  c(a  +  &)• 

(2) 


(a  +  c)  («2  +  c2)  _  (6  +  d)  (b2  +  d2) 


(a  -  c)  (a2  -  c2)       (6  -  d)  (b2  -  cZ2) 

,„.   pa2  +  g«&  +  ?'ft2   _  pc2  +  QCf?  +  rd2 
la2  +  m«6  +  nb2        lc2  +  ma?  +  nd2' 

26.  If  a;  and  y  be  unequal  and  x  have  to  ?/  the  duplicate 
ratio  of  x  +  z  :  y  +  2,  prove  that  2  is  a  mean  proportional 
between  x  and  y. 

27.  If  a:  fr  ::»:?,  prove  that  a2 -f&2:  -^—  ::p2+g2:  -^— . 

a  +  &  p+g 

28.  If  four  quantities  are  in  proportion,  and  the  second  is 
a  mean  proportional  between  the  third  and  fourth,  prove 
that  the  third  will  be  a  mean  proportional  between  the  first 
and  second. 

29  If a  +  h  +  c  +  d  =  a  +  b-c-d^ prove that 

a  —  b  +  c  —  d       a  —  b  —  c  -\-  d 
6,  c,  d  are  in  proportion. 

30.  Each  of  two  vessels  contains  a  mixture  of  wine  and 
water ;  a  mixture  consisting  of  equal  measures  from  the  two 
vessels  contains  as  much  wine  as  water,  and  another  mixture 
consisting  of  four  measures  from  the  first  vessel  and  one 
from  the  second  is  composed  of  wine  and  water  in  the  ratio 
of  2:3.  Find  the  proportion  of  wine  and  water  in  each  of 
the  vessels. 

Ans.  In  the  first  the  wine  is  J  of  the  whole ;  in  the 
second  §. 


EXAMPLES.  317 

31/   If  x  oc  ?/,  and  y  =  7  when  a;  =  18,  find  x  when  y  =  21. 

-i//.v.  54. 

32y'lf  x  oc  -,  and  y  =  4  when  as  =  1 5,  find  y  when  a;  =  6. 
y  An*.  10. 

33.  J.  varies  jointly  as  7?  and  C;  and  ^1  =  6  when  5  =  3 
and  C  =  2 :  find  ^L  when  5  =  5  and  C  =  7.  4w*.  35. 

34.  A  varies  jointly  as  5  and  C;  and  A  =  9  when  5  =  5 
and  C  =  7  :  find  £  when  A  =  54  and  C  =  10.       Am.  21. 

35.  A  varies  directly  as  5  and  inversely  as  C ;  and  ^4=10 
when  5=15  and  C  =  6  :  find  A  when  5  =  8  and  C  =  2. 

Ans.  16. 

36.  If  3rt  +  76  oc  3a  +  136,  and  a  =  5  when  6  =  3.  find 
the  equation  between  a  and  6.  Ans.  3«  =  56. 

37.  .1  oc  5,  and  A  =  2  when  B  =  1  ;  find  ^1  when  5  =  2. 

Jj>s.  4. 

38.  If  A2  +  52  cc  ^42  -  52,  prove  that  A  +  B  cc  A  -  B. 

39.  3  J.  +  55  oc  5A  4-  35  ;  and  .4  =  5  when  5  =  2:  find 
the  ratio  A:  B.  Ans.  5:2. 

40.  A  oc  »5  +  C;  and  .4  =  4  when  5  =  1  and  C  =  2  ; 
and  J.  =  7  when  5=2  and  (7  =  3:  find  ».  ^4??6\  2. 

41.  If  a?2  cc  ?/3,  and  a;  =  2  when  ?/  =  3,  find  the  equation 
between  a;  and  y.  Ans.  27x*  =  4y3. 

42.  If  ?/  varies  as  the  sum  of  two  quantities,  one  of  which 
varies  as  x  directly,  the  other  as  x  inversely,  and  if.  y  =  4 
when  x  =  1.  and  ?/  =  5  when  a;  =  2,  find  the  equation 
between  a;  and  y.  ^       =  ^  +  2 

43.  If  ?/  =  the  sum  of  two  quantities,  one  of  which  varies 
directly  as  a*,  and  the  other  inversely  as  SB2;  and  if  y  =  19 
when  a?  =  2,  or  3  ;  find  y  in  terms  of  a?.     A   _       _  r„   ,    36 

J  X1 

44.  If  ?/  varies  as  the  sum  of  three  quantities  of  which 
the  first  is  constant,  the  second  varies  as  a;,  and  the  third  as 
x2;  and  if  y  =  0  when  x  =  1 ,  ?/  =  1  when  a;  =  2,  and 
y  =  4  when  a;  =  3  ;  find  ^/  when  a;  —  7.  Ans.  36. 


318  DEFINITIONS  —  FORM  UL^E. 


CHAPTER   XVL 

ARITHMETIC,     GEOMETRIC,     AND     HARMONIC 
PROGRESSIONS. 

ARITHMETIC    PROGRESSION. 

157.  Definitions  —  Formulae.  —  A  number  of  terms 
formed  according  to  some  law  is  called  a  series.  Quantities 
are  said  to  be  in  Arithmetic  Progression*  when  they  increase 
or  decrease  by  a  constant  difference,  called  the  common 
difference. 

Thus,  the  following  series  are  each  in  Arithmetic  Progres- 
sion : 

2,5,8,  11,  14,  17, 

J,  /,  o,  «j,  i,  — i,  — o,  — o,  •■••••■ 

a,  a  -f-  d,  a  +  2d,  a  +  3d,  a  -f  4c?, 

The  letters  A.  P.  are  often  used  for  shortness  instead  of 
the  term  Arithmetic  Progression. 

The  common  difference  is  found  by  subtracting  any  term 
of  the  series  from  that  which  immediately  follows  it.  In  the 
first  series  above  the  common  difference  is  3  ;  in  the  second 
it  is  —  2  ;  in  the  third  it  is  d. 

The  series  is  said  to  be  increasing  or  decreasing,  according 
as  the  common  difference  is  positive  or  negative.  Thus,  the 
first  series  above  is  increasing,  and  the  second  is  decreasing. 
If  we  examine  the  third  series  above,  we  see  that  the  coef- 
ficient of  d  in  any  term  is  less  by  one  than  the  number  of  the 
term  in  the  series. 

Thus  the  2d   term  is  a  -f  d, 

3d    term  is  a  +  2d, 
4th  term  is  a  +  3d, 

*  Called  liliso  Arithmetic  /Series. 


DEFINITIONS  —  FORMULAE.  319 

and  so  on.     Hence  if  n  be  the  number  of   terms,  and  if  I 
denote  the  last,  or  »th  term,  we  have 

I  =  a  +  (n  -  \)d (1) 

Let  S  denote  the  sum  of  n  terms  of  this  series ;  then  we 
have 

S  =  a  +  (a  +  d)  +  (a  +  2d)  +...+(/-  2d)  +  (l  -  d)  +J ; 
and,  by  writing  the  series  in  the  reverse  order,  we  have 
S  =  I  +(l  -  d)  +  (l  -  2d)  +  .  .  .  +(a  +  2d)  +  (a  +  d)  +  a. 
Adding  together  these  two  equations,  we  have 

2S  =  (a  +  0  +  (a  +  I)  +  (a  +  I)  + to  n  terms 

=  n{a  +  0- 
.-•     ^  =  |(a  +  i) (2) 

By  (1)  and  (2)  we  have  S  =  ^[2a  +  (n  -  l)d]     .     .   (3) 

"We  have  here  three  useful  formulae,  (1),  (2),  (3),  which 
should  be  remembered;  in  each  of  these  any  one t of  the 
letters  may  denote  the  unknown  quantity  when  the  other 
three  are  known.  For  example,  in  (1),  we  can  write  down 
any  term  of  an  A.  P.  when  the  first  term,  the  common  dif- 
ference, and  the  number  of  the  term  are  given.  Thus,  if  the 
first  term  of  an  A.  P.  is  5  and  the  common  difference  is  3, 

the  10th  term  =  5  +  (10  -  1)3  =  32, 

and  the  20th  term  =  5  +  (20  -  1)3  =  62. 

Also  in  (2),  if  we  substitute  given  values  for  S,  n,  I,  we 
obtain  an  equation  for  finding  a;  and  similarly  in  (3). 
Thus, 

1.  Find  the  sum  of  20  terms  of  the  series  1,  3,  5,  7,  .  .  . 

Here  a  =  1,  d  =  2,  ?i  =  20  ;  therefore  by  (3) 

S  =  -22°-[2  +  19  X  2]  =  10  X  40  =  400. 

2.  The  first  term  of  a  series  is  5,  the  last  45.  and  the  sum 
400  ;  find  the  number  of  terms,  and  the  common  difference. 


320  DEFINITIONS  —  FORM  VLJE. 

Here        a  =  5,  I  =  45,  S  =  400  ;  therefore  by  (2) 
400  =  -(5  -f-  45)  =  25?*.         .-.     n  =  16. 

By  (1)  45  =  5  +  15d.         .-.     d  =  2§. 

When  any  two  terms  of  an  A.  P.  are  given,  the  series  can 
be  completely  determined ;  for  the  data  furnish  two  simul- 
taneous equations,  with  two  unknown  quantities,  which  may 
be  solved  by  methods  previously  given. 

3.  The  10th  and  15th  terms  of  an  A.  P.  are  25  and  5 
respectively  ;  find  the  series. 

Here  25  =  a  +  9cZ ; 

and  5  =  a  -f  14(7. 

By  subtraction,  20  =  —  bd.         .-.     d  =  —4. 

Then  a  =  5  -  14cZ  =  61. 

Hence  the  series  is  61,  57,  53, 

4.  Find  the  sum  of  the  first  n  odd  integers. 
Here  a  =  1,  and  d  =  2  ;  therefore  by  (3) 

*     £  =  ?^[2  +  (»  ~  1)2]  =  ^  X  2,i  =  n\ 

Thus  the  sum  of  any  number  of  consecutive  odd  integers 
beginning  with  unity,  is  the  square  of  their  number.* 
Find  the  last  term  and  sum  of  the  following  series : 

5.  14,  64,  114, to  20  terms.  Ans.  964,  9780. 

6.  9,  5,  1, to  100  terms.  -387,  -18900. 

7.  h  ~b  ~b to  21  terms.  -9|,  -99f. 

Fiud  the  sum  of  the  following  series : 

8.  5,  9,  13, to  19  terms.  779. 

9.  12,  9,  6, to  23  terms.  -483. 

Find  the  series  in  which 

10.    The  27th  term  is  186,  and  the  45th  is  312. 

Ans.  4,  11,  18, 

♦Thin  proposition  WM  known  to  the  Greek  geometers. 


ARITHMETIC  MEAN.  321 

11.  The  9th  term  is  -11,  and  the  102d  is  -150 J. 

Ans.  1,  —  £,  —  2, 

12.  The  16th  term  is  214,  and  the  51st  is  739. 

Ans.  —11,  4,  19, 

158.  Arithmetic  Mean.  —  When  three  quantities  aro  in 
Arithmetic  Progression,  the  middle  one  is  called  the  Arith- 
metic Mean  of  the  other  two. 

Thus  if  «,  6,  c  are  in  A.  P.,  b  is  the  arithmetic  mean  of  a 
and  c ;  and  by  the  definition  of  A.  P.  we  have 

b  —  a  =  c  —  b; 
...     h  =  l(a  +  C). 

Thus  the  arithmetic  mean  of  any  two  quantities  is  half  their 
sum. 

Between  any  two  given  quantities  any  number  of  terms 
may  be  inserted  so  that  the  whole  series  thus  formed  shall 
be  in  A.  P.  ;  the  terms  thus  inserted  are  called  the  arithmetic 
means. 

For  example,  to  insert  four  arithmetic  means  between  10 
and  25. 

Here  we  have  to  find  an  A.  P.  with  4  terms  between  10  and 
25,  so  that  10  is  the  first  and  25  is  the  sixth  term. 

By  (1)  of  Art.  157, 

25  =  10  +  M;         .-.     d  =  3. 

Thus  the  series  is  10,  13,  16,  19,  22,  25  ; 
and  the  required  arithmetic  means  between  10  and  25  are 

13,  16,  19,  22. 

In  general.     To  insert  n  arithmetic  means  between  a  and  b. 
Here  we  have  to  find  an  A.  P.  with  n  terms  between  a  and 
6,  so  that  a  is  the  first  and  b  is  the  (n  -f-  2)th  term. 
By  (1)  of  Art.  161, 

b  =  a  +  (n  +  2  -  \)d  =  a  +  (n  +  l)d 

,       b  —  a 

.-.     d  =  -. 

n  +  1 


322  ARITHMETIC   MEAN. 

Thus  the  required  means  are 

b  —  a  06  —  a  „    .     b  —  a 

n  4-  1  n  +  1  n  4-  1 

1.  Find  the  sum  of  the  first  p  terms  of  the  series  whose 
nth  term  is  Sn  —  1. 

By  putting  n  =  1,  and  n  =  p  respectively,  we  obtain 

first  term  =  2,  last  term  =  Sp  —  1. 

Hence  by  (2),  Art.  157,  S  =  £(2  +  3p  -  1)  =  £(3p  +  l). 

In  an  Arithmetic  Progression  when  a,  #,  and  d  are  given, 
n  is  to  be  found  by  solving  the  quadratic  (3),  Art.  157. 
When  both  roots  are  positive  and  integral,  there  is  no  diffi- 
culty in  interpreting  the  result  corresponding  to  each. 

2.  How  many  terms  of  the  series  24,  20,  16, must 

be  taken  that  the  sum  may  be  72  ? 

Here  a  =  24,  d  =  -4,  S  =  72.  Then  from  (3),  Art. 
157,  we  have 

72  =  -[2  x  24  +  (n  -  l)(-4)]  =  24?i  -  2n(n  -  1). 

Ld 

.-.  ' V  -  13m  +  36  =  0,  or  (n  -  4)(n  -  9)  =  0. 
,\     7i  =  4,  or  9. 

Both  of  these  values  satisfy  the  conditions  of  the  ques- 
tion ;  for  if  we  take  the  first  4  terms,  we  get  24,  20,  16,  12  ; 
and  if  we  take  the  first  9  terms,  we  get  24,  20,  16,  12,  8,  4, 
0,  —4,  —8,  in  either  of  which  the  sum  is  72;  the  last  5 
terms  of  the  last  series  destroy  each  other,  so  that  the  sum 
of  the  first  4  terms  is  the  same  as  the  sum  of  the  first  9 
terms. 

When  one  of  the  roots  is  negative  or  fractional,  it  is  inap- 
plicable, for  a  negative  or  a  fractional  number  of  terms  is, 
strictly  speaking,  without  meaning.  In  some  cases  however 
a  suitable  interpretation  can  be  given  for  a  negative  value 
of  71. 


ARITHMETIC  MEAN.  323 


3.  How  many  terms  of  the  scries  —0,-6,-3, 

must  be  taken  that  the  sum  may  be  66  ? 

Here  66  =  '-[-18  +  (n  -  1)3]. 

.-.     n"  —  In  —  44  =  0;     or    (n  -  11)  (n  +  4)  =  0. 
.-.     n  =  11,  or  —4. 

If  we  take  11  terms  of  the  series,  we  have 

-9,  -6,  -3,  0,  3,  6,  9,  12,  15,  18,  21  ; 

the  sum  of  which  is  QQ. 

If  we  begin  at  the  last  term  and  count  backwards  four 
terms,  the  sum  is  also  66.  From  this  we  see  that,  although 
the  negative  solution  does  not  directly  answer  the  question 
proposed,  we  are  enabled  to  give  it  an  intelligible  meaning 
as  follows :  begin  at  the  last  term  of  the  series  which  is 
furnished  by  the  positive  value  of  »,  and  count  backwards  for 
as  many  terms  as  the  negative  value  indicates  ;  then  the 
result  will  be  the  given  sum.  We  thus  see  that  the  negative 
value  for  n  answers  a  question  closely  connected  with  that 
to  which  the  positive  value  applies. 

4.  How  many  terms  of   the   series  26,  21,   16, 

must  be  taken  that  the  sum  may  be  74? 

Here  74  =  |[52  +  (n  -  l)(-5)]. 

Solving,  we  get  n  =  4,  or  7j. 

Thus,  the  only  applicable  value  of  n  is  4.  We  infer  that 
of  the  two  numbers  7  and  8,  one  corresponds  to  a  sum 
greater,  and  the  other  to  a  sum  less  than  74. 

5.  Insert  3  arithmetic  means  between  12  and  20. 

Arts.  14,  16,  18. 

6.  Insert  5  arithmetic  means  between  14  and  16. 

Ans.  14$,  14f 

7.  Insert  17  arithmetic  means  between  93  and  69. 

Ans.  91|,  90$, 70$. 


324  GEOMETRIC  PROGRESSION. 

How  many  terms  must  be  taken  of  the  series 

8.  42,  39,  36, to  make  315?      Ans.  14,  or  15. 

9.  -16,-15,-14, to  make -100?       8,  or  25. 

10.  20,  18},  17J, to  make  162±?  13,  or  20. 

11.  The  sum  of  three  numbers  in  A.  P.  is  39,  and  their 
product  is  2184  ;  find  them.  Ans.  12,  13,  14. 

12.  The  sum  of  10  terms  of  an  A.  P.,  whose  first  term 
is  2,  is  155  ;  what  is  the  common  difference?  Ans.  3. 

GEOMETRIC    PROGRESSION! 

159.  Definition  —  Formulae.  —  Quantities  are  said  to 
be  in  Geometric  Progression  when  they  increase  or  decrease 
by  a  constant  factor,  called  the  common  ratio. 

Thus,  the  following  series  are  each  in  Geometric  Progres- 
sion (G.  P.): 

3,  6,  12,  24,  48, 

Q      1  111 

°»    X»      "3'       9'    27' 

a,  a?%  «r2,  a/*3,  or4, 

The  common  ratio  is  found  by  dividing  any  term  of  the 
series  by  that  which  immediately  precedes  it.  In  the  first 
series  above  the  common  ratio  is  2  ;  in  the  second  it  is  ^  ;  in 
the  third  it  is  r. 

The  series  is  said  to  be  increasing  or  decreasing,  according 
as  the  common  ratio  is  greater  than  1,  or  less  than  1. 
Thus,  the  first  series  above  is  increasing,  and  the  second  is 
decreasing. 

Note  1.  — An  Arithmetic  Progression  is  formed  by  repeated  addition 
or  subtraction;  a  Geometric  Progression  by  repeated  multiplication  or 
division. 

If  we  examine  the  third  series  above,  we  see  that  the 
exponent  of  r  in  any  term  is  less  by  one  than  the  number  of 
the  term  in  the  series. 

Thus,  the  2d   term  is  ar, 

3d    term  is  ar2, 
4th  term  is  ar3, 


GEOMETRIC  PROGRESSION.  325 

and  so  on.     Hence  if  n  be  the  number  of  terms,  and  if  I 
denote  the  last,  or  ?<lh  term,  we  have 

I  =  or"-1 (1) 

Let  S  denote  the  sum  of  n  terms  of  this  series  ;  then  we 
have 

S  =  a  +  ar  +  ar2  + +  cu"-2  +  at*'1; 

multiplying  by  r,  we  have 

Sr  =  ar  +  ar2  + +  «/-n"2  +  a/-71-1  +  ar9. 

Hence  by  subtraction,  we  have 

Sr  —  S  =  «/•"  -  «  ;  _  or    (r  -  1)6'  =  a(r"  -  1). 

^     ^  =  a(/-  -  1}>  Qr  «(1  -  ^) (2) 

r  —  1  1  —  >' 

Multiplying  (l)~by  r,  and  substituting  in  (2),  we  get 

0       rl  —  a         a  —  vl  /ox 

S  = -,  or -,      ....   (3) 

T   —    1  1    —    T 

a  form  which  is  sometimes  useful. 

Xote  2.  —  It  will  be  found  convenient  to  remember  both  forms 
given  in  (2)  for  S,  and  to  use  the  first  form  in  all  cases  when  r  is 
positive  and  >  1,  and  the  second  when  r  is  negative  or  <  1. 

1.    Find  the  8th  term  of  the  series  —  J,  J,  —  f , 

Here  a  =  —  -J.  n  =  8,  r  =  \  -h  (  —  -J)  =  — |:    therefore 


by  (i) 


7    1/        S\>    1  ( 2187\ 

1    —    —2\~2)      —    ~3\ T38V 

=  xf§  =  the  8th  term- 


2.  Sum  the  series  1,  3,  9, to  6  terms. 

Here  a  =  1,  n  =  6,  r  —  3  ;    therefore  by  the  first  form 
of  (2),  =  #_!  =  720  -  1  =  364> 

3-1  2 

3.  Sum  the  series  81,  54,  36, to  0  terms. 

Here  a  =  81,  n  =  9,  r  =  54  -=-  81  =  |  ;  therefore  by  the 
second  form  of  (2), 

S  =  81C1  ~  (jQ  =  243[1  -  (|)9] 
1  ~~  I 

—    9  1Q   512,   _    9?,fi55 

+  °  8  1      ""    - ,JU8  1* 


326  GEOMETRIC  MEAN. 

4.  Sum  the  series  2,  —  3,  f ,  — to  7  terms. 

Here  a  =  2,  n  =  7,  r  =  —  | ;  therefore  by  the  second  form 

°f(2)'         ^_2fl  -(-|)71  _2fl  +  \W1 
l-(-f)  I 

=  I   X   Ws5  =  14H- 

5.  Find  the  6th  term  of  each  of  the  following  series : 

(1)   9,  3,  1,  etc.  ;     (2)   2,  -3,  |,  etc.  ;     (3)  d\  ab,  b2,  etc. 

Ans.  (1)  J?;     (2)  -W;     (8)  £ 

ft 

Sum  the  following  series  : 

6.  1,      4,  1G, to  6  terms.  -4ns.  1365. 

7.  25,     10,    4, to  4  terms.  40§ . 

8.  |,  -1,    |, to  7  terms.  ^-. 

9.  3,  -1,    i, to  6  terms.  2-§£. 

160.   Geometric  Mean.  —  When  three  quantities  are  in 

Geometric  Progression  the  middle  one  is  called  the  Geometric 
Mean  between  the  other  two. 

Thus  if  «,  6,  c  are  in  G.  P.,  b  is  the  geometric  mean 
between  a  and  c  ;  and  by  the  definition  of  G.  P.,  we  have 
b  _  c. 
ft       6' 
.-.     b2  =  ftc;         .-.     6  =  V^c. 

Thus,  £/*e  geometric  mean  between  any  two  quantities  is  the 
square  root  of  their  product. 

Quantities  which  are  in  G.  P.  are  in  continued  proportion, 
and  the  geometric  mean  between  two  quantities  is  the  same 
as  their  mean  proportional  (Art.  152). 

Between  any  two  given  quantities  any  number  of  terms  may 
be  inserted  so  that  the  whole  series  thus  formed  shall  be  in 
G.  P.  ;  the  terms  thus  inserted  are  called  the  geometric  means. 

For  example,  to  insert  three  geometric  means  between  2 
and  32. 

Here  we  have  to  find  a  G.  P.  with  3  terms  between  2  and 
32,  so  that  2  is  the  first  and  32  is  the  fifth  term. 

By  (1)  of  Art.  159,  32  =  2/-4 ;         .-.     r  =  2, 


THE  SUM    OF   AN   INFINITE  NUMBER   OF   TERMS    327 

Thus  the  series  is  2,  4,  8,  16,  32,  and  the  required 
geometric  means  between  2  and  32  are  4,  8,  16. 

In  general.      To  insert  n  geometric  means  between  a  and  b. 

Here  we  have  to  find  a  G.  P.  with  n  terms  between  a  and 
b,  so  that  a  is  the  first  and  b  is  the  (n  -f-  2)th  term. 

By  (1)  of  Art.  159, 

b  =  or**1:         .-.     r"+1  =  -: 


V  a 


(1) 


Ans.  -28,  14,  -7,  1 


Thus  the  required  means  are  ar,  ar2, ar",  where  r 

lias  the  value  found  in  (1). 

1.    Insert  4  geometric  means  between  160  and  5. 

Ans.  80,  40,  20,  10. 

7 

_7      7 
4'    8* 

3.    Insert  4  geometric  means  between  5J  and  40J. 

Ans.  8,  12,  18,  27. 
161.  The  Sum  of  an  Infinite  Number  of  Terms.  — 
From  (2)  of  Art.  159,  we  have 

s  =  a(l  -  r")  =  _^ ^L  0) 

1  -  r  1  -  r        1  -  r 

Now  suppose  r  is  a  proper  fraction,  positive  or  negative; 

then  the  greater  the  value  of  n  the  smaller  is  the  absolute 

ft/*71 
value  of  ?•",  and    consequently  of  ;    and  by  taking  n 

sufficiently  large  rn  can    be   made   as   small   as   ice   please. 
Hence,  by  taking  n  large  enough,  the  sum  of  n  terms  of  the 

series  can  be  made  to  differ  from by  as  small  a  quan- 

1  —  r 

tity  as  we  please. 

Thus,  the  sum  of  an  infinite  number  of  terms  of  a  decreas- 
ing Geometric  Progression  is ; 

or  more  brief!}',  the  sum  to  infinity  is . 


328   THE  SUM   OF  AN  INFINITE  NUMBER   OF  TERMS. 


This  quantity,  ,  which  we  call  the  sum  of  the  series, 

is  the  limit  to  which  the  sum  approaches,  but  never  actually 
attains ;  that  is,  although  no  definite  number  of  terms  will 

amount  to ,  yet  by  taking  a  sufficient  number,  the  sum 

1  —  r 

will  reach  it  as  near  as  we  please. 

1.    Sum  the  series  -J,  J,  J, 

For  n  terms  we  have  by  (2)  of  Art.  159, 


S 


=  V !v  =  i  _  1. 


i-i 


From  this  result  it  appears  that  however  many  terms  be 
taken,  the  sum  of  this  series  is  always  less  than  1.     Also  we 

see  that  by  taking  ?i  large  enough,  the  fraction  —  can  be 

made  as  small  as  we  please.  Hence  by  taking  a  sufficient 
number  of  terms,  the  sum  can  be  made  to  differ  from  1  by 
as  little  as  we  please ;  and  when  n  is  made  infinitely  great  we 
have  8  =  1. 

This  may  be  illustrated  geometrically  as  follows: 

A\ 1 1 1 1 \B 

Px  P*  Ps    A 

Let  AB  be  a  line  of  unit  length.  Bisect  AB  in  Px ;  bisect  PXB  in 
P2,  P2B  in  P3,  P3B  in  P4,  and  so  on  indefinitely,  always  bisecting 
the  remaining  distance.  It  is  evident  that  by  a  series  of  such  bisections 
we  can  never  reach  B,  because  we  shall  always  have  a  distance  left 
equal  to  half  the  preceding  distance;  but  by  a  sufficient  number  of 
these  bisections  we  can  come  nearer  to  B  than  any  assigned  distance, 
however  small,  because  every  bisection  carries  us  over  half  the  remain- 
ing distance.  That  is,  if  we  take  a  sufficient  number  of  terms  of  the 
series         Ap^  +  p^  +  p^  +  p^  + ? 


we  shall  have  a  result  differing  from  AB,  i.e.,  from  unity,  by  as  little 
as  we  please.     This  is  simply  a  geometric  way  of  saying  that 

1  +  I  +  1  + to  oo  =  1. 

2  t  2a  I"  2»  -r 


VALUE   OF  A   REPEATING   DECIMAL.  329 

Sum  the  following  series  to  inanity : 
2.  1,^,1 Am.  2. 

3-  9,  -6,  4,  - 5f 

4.  1,  -i,  J, f. 

5-   1,  1,  A, ft. 

162.  Value  of  a  Repeating*  Decimal.— "Repeating 
decimals  furnish  a  good  illustration  of  inlinite  Geometric 
Progressions. 

1.  Find  the  value  of  .423. 

.423  =  .4232323  

=  A  +  ^  +  !3         

10        108       10s 

=  ±+  ?8/i+_L+JL+  \ 

io     io3v       io2     io4      / 

102/ 

=   4         23        100        4         23    =  419 
10        103         99         10        990  ~  990' 

which  agrees  with  the  value  found   by  the   usual   rule   in 
Arithmetic. 

The  value  of  any  repeating  decimal  may  be  found  by  the  method 
employed  in  the  last  example;  but  in  practice  it  maybe  found  more 
easily  by  a  general  rule,  which  may  be  proved  as  follows : 

Let  P  denote  the  figures  which  do  not  repeat,  and  suppose  them  p 
in  number;  let  Q  denote  the  repeating  period  consisting  of  q  figures. 
Let  S  denote  the  value  of  the  repeating  decimal ;  then 

8  =  .PQQQ ; 

.-.     IO?*  =  P.QQQ ; 

and  10P+QS  =  PQ.QQQ ; 

♦Called  also  recurring  and  circulating. 


=  Io  +  U^— "^ 


330 


HARMONIC  PROGRESSION. 


by  subtracting, 

(IOp+7  —  10p)S  =  PQ—  P; 
that  is,  10^(10?  -  1)S  =  PQ  -  P; 

PQ  -  P 


S  - 


(10« 


1)10" 


Now  10^  —  1  is  a  number  consisting  of  q  nines;  therefore  the 
denominator  consists  of  q  nines  followed  by  p  ciphers.  Hence,  for 
finding  the  value  of  a  repeating  decimal,  we  have  the  following 

Kule.  Subtract  the  integral  number  consisting  of  the  non-repeating 
figures  from  the  integral  number  consisting  of  the  non-repeating  and 
repeating  figures,  and  divide  by  a  number  consisting  of  as  many  nines 
as  there  are  repeating  figures  followed  by  as  many  ciphers  as  there  are 
non-repeating  figures. 

Find  the  value  of  the  following  repeating  decimals  : 


2.    .151515  ....   Ans.  ^, 


3.    .123123123  .  .  . 


JUL 
333' 


4.  .16  . 

5.  .037. 


HARMONIC    PROGRESSION. 

163.  Definition.  —  A  series  of  quantities  is  said  to  be  in 
Harmonic  Progression  when  their  reciprocals  are  in  Arith- 
metic Progression. 

Thus,  the  following  series 

fill  anA    1212 

1l   "3'  "5>    7' anU   4'    71    31   p » 

are  each  in  Harmonic  Progression   (II.  P.)  because   then 
reciprocals, 

1,  3,  5,  7, and  4,  3J,  3,  2£,  .  .  .  .  .  . 

are  in  A.  P. 

The  following  is  therefore  a  general  form  for  a  H.  P0 

1        1  1  1 

a  a  +  d   a  +  2d  a  +  (n  —  l)d 

because  the  reciprocals  of  the  terms  are  in  A.  P. 

From  the  above  definition  it  follows  that  all  problems 
relating  to  quantities  in  II.  P.  can  be  solved  by  taking  the 


HARMONIC   MEAN.  331 

reciprocals  of  the  quantities  and  using  the  formula?  relating 
to  A.  P.  This  makes  it  unnecessary  to  give  any  special 
rules  for  the  solution  of  problems  in  H.  P. 

If  a,  6,  c  be  three  consecutive  terms  of  a  II.  P.,  then  we 
have  by  definition, 

1  _  1  =  1  _  1 

b       a        c        b 

a  —  5  _  5  —  c 

ab  be 

.'.     a  —  b  :b  —  c: :  a:  c       .     .     .     .    (1) 

Thus,  if  three  quantities  are  in  Harmonic  Progression,  the 
difference  between  the  first  and  the  second  is  to  the  difference 
between  the  second  and  the  third  as  the  first  is  to  the  third. 

Sometimes  this  relation  is  taken  as  the  definition  of  Har- 
monic Progression. 

1.  The  12th  term  of  a  H.  P.  is  J,  and  the  19th  term  is 
-j32  ;  find  the  series. 

Here  the  12th  and  19th  terms  of  the  corresponding  A.  P. 
are  5  and  %2-  respectively.     Therefore  by 

(1),  Art.  161,  5    =  a  +  lid, 

and  -232  =  a  +  I8d. 

Solving,  we  get  d  =  },  a  =  f . 

Hence  the  A.  P.  is  f ,  f ,  2,  f ,  { 

and  the  H.  P.  is      hbhhh 

Find  the  last  term  of  the  following  harmonic  series : 

2.  4,      2,    1|, to    G  terms.  Ans.  -§. 

3.  2J,  lif,  1T^, to  21  terms.  &• 

4.  1$,  lfi,  2T23, to    8  terms.  -4. 

164.  Harmonic  Mean.  —  When  three  quantities  are  in 
Harmonic  Progression,  the  middle  one  is  called  the  Harmonic 
Mean  between  the  other  two. 


332  HARMONIC  MEAN. 

Thus  if  «,  6,  c  are  in  H.  P.,  b  is  the  harmonic  mean 
between  a  and  c ;  and  by  the  definition  of  H.  P.  we  have 

1  _  1  =  1  _  1. 
b       a       c       b 

•■■  hi  +  1- 

b       a       c 

,  2ac 

.*.     b  = 


a  -\-  c 


Thus,  the  harmonic  mean  between  any  two  quantities  is 
twice  their  product  divided  by  their  sum. 

Between  any  two  given  quantities  any  number  of  terms 
may  be  inserted  so  that  the  whole  series  thus  formed  shall 
be  in  H.  P.  ;  the  terms  thus  inserted  are  called  the  harmonic 
means. 

For  example,  to  insert  5  harmonic  means  between  -§  and 

A- 

Here  we  have  to  insert  5  arithmetic  means  between  | 
and  Jjf .     Hence,  by  (1)  of  Art.  157, 

1 5    _    3      i     ar] .  .         r7   _     1 

-g"    —    "2      i      Da  »  *    *       u    —    TO"* 

Thnsj  flip    A      P     i<s   3      2  5      26      27      28      29      15. 

and  therefore  the  required  harmonic  means  between  f  and  ^ 

«rp  1616161616 

clie  "25'  "215"'    27'    28'  "29* 

In  general.     To  insert  n  harmonic  means  between  a  and  b. 
Here  we  have  to  insert  n  arithmetic  means  between  -   and 


a 


i.     By  Art.  158,  these  will  be 

11  1  _  1 

1    .    b       a    1    ,    9  b       a 


+  :— — '  ~  +  2 


a       n  -\-  1    a  n  +  1 

\)b  +  (a  -  fr)    (n  +  1 
(ji  -f-  l)ab  {n  H-  1)«6 


that  is,  {n  +  1)?)  +  (a  ~  ?v)    (*-H-l)ft  +  2(a-5) 


RELATION   BETWEEN   THE    THREE   MEANS.  333 

and  therefore  the  required  harmonic  means  between  a  and  b 
are  the  reciprocals  of  these,  that  is, 

(n  +  I)  (ib                        (n  +  l)ab 
(n  +  1)6  +  (a  -  by  (n  +  1)6  +  2(a  -  6)' 

1.  Insert  2  harmonic  means  between  4  and  2.     ^4?is.  3,  Jg2. 

2.  Insert  3  harmonic  means  between -J  and  ^T.     ^,  ^,  -^. 

3.  Insert  4  harmonic  means  between  1  and  G.     1  J,  lj,  2,  3. 

165.  Relation  between  Arithmetic,  Geometric,  and 
Harmonic  Means. 

(1)  If  A,  G,  H  be  the  arithmetic,  geometric,  and  har- 
monic means  between  a  and  6,  then  (Arts.  158,  160,  164), 

A  =  <±±± (1) 

2  v  ' 

G  =  </ab (2) 

#=^V (3) 

a  +  b 

Therefore    AH  =  °^±^  X  -^-  =  ab  =  £2; 
2  a  +  6 

that  is,  G  is  the  geometric  mean  between  A  and  H. 

Hence  the  geometric  mean  between  any  two  real  positive 
quantities,  a  and  6,  is  also  the  geometric  mean  between  the 
arithmetic  and  the  harmonic  means  between  a  and  b. 

(2)  From  (1)  and  (2)  we  have 

A  _  G  =  <±st±  -  ^  =  i(v/-  -  ^)2; 
and  from  (2)  and  (3), 

a  +  b       a    f  b 
Now  if  a  and  b  are  both  positive,  s/a  and  s/b  are  both  real ; 
therefore   (Va  —  V^)2  is  positive  ;  also  \fab  and  a  +  b  are 
both   positive.     Hence  A  —  G   and    G  —  II  are   positive. 
Therefore  -4  >  C?  >  if. 


334  EXAMPLES. 

That  is,  the  arithmetic,  geometric,  and  harmonic  means 
between  any  two  real  positive  quantities  are  in  descending 
order  of  magnitude.* 

Three  quantities,  a,  b,  C,  are  in  A.  P.,  G.  P.,  or  H.  P., 
according  as 

a_j— —  _  a^   a^  Qr  a^  respectively. 
b  —  c       a     b         c 

The  first  follows  from  the  definition  of  A.  P.  (Art.  157). 
In   the   second,  6 (a  —  b)  =  a(b  —  c)  ;         .*.     b2  =  ac. 
See  Art.  160. 

The  third  follows  from  (1)  of  Art.  163. 

Harmonic  properties  are  interesting  chiefly  because  of  their  im- 
portance in  Geometry  and  in  the  Theory  of  Sound.  If  there  be  a 
series  of  strings  of  the  same  substance,  the  lengths  of  which  are 
proportional  to  1,  ?,  %,  \,  £,  &,  and  if  these  strings  are  stretched  tight 
with  equal  force,  and  any  two  of  them  are  sounded  together,  the  effect 
is  found  to  be  harmonious  to  the  ear. 

Notwithstanding  the  comparative  simplicity  of  the  law  of 
its  formation,  there  is  no  general  formula  for  the  sum  of  any 
number  of  terms  in  harmonic  progression. 

EXAMPLES. 

Find  the  last  term  and  sum  of  the  following  series : 

1.  1,  1.2,  1.4, to  12  terms.  Ans.  3.2,  25.2. 

2.  3|,  1,  -1J, to  19  terms.  —41$,  -361. 

3.  64,  96,  128, to  16  terms.  544,  4864. 

Sum  the  following  series  : 

4.  4,  5J,  6J, to  37  terms.  980J. 

5.  -3,  1,  5, to  17  terms.  403. 

6.  3a,  a,  —a, to  a  terms.  a2(4  —  a). 

7.  H,  **i  *h to  n  terms.  5"(°12~  ?0* 

*  These  two  proposition*  were  known  to  the  Greek  geometers. 


EXAMPLES.  335 

8.  1±,  m,  2§f, to  n  terms.        Ans.  n(17  +  7n). 

9.  —J ,  V^,  —J ,  ....  to  7  terms.       7(^2  +  2). 

&  +  1  v/2  -  1 

Find  the  series  iu  which 

10.  The  15th  term  is  25,  and  the  29th  term  46. 

Ans.  4,  5J,  7, 

11.  The  15th  term  is  -25,  and  the  23d  term  -41. 

Ans.  3,  1,  —1, 

12.  Insert  14  arithmetic  means  between  —  7J  and  —  21. 

Ans.  -6«,  -6&, -2TV 

13.  Insert  36  arithmetic  means  between  8J  and  2£. 

4**.  8J,  8J, 2§. 

How  manj'  terms  must  be  taken  of  the  series 

14.  15|,  15J,  15, to  make  129?     Ans.  9,  or  86. 

15.  — 10|,  -9,  -7J,  ...  to  make  -42?  7,  or  8. 

16.  -6§,  -6f,  -6, ....  to  make  -524 ?         11,  or  24. 

17.  The  sum  of  three  numbers  in  A.  P.  is  33,  and  their 
product  is  792  :  find  them.  Ans.  4,  11,  18. 

18.  An  A.  P.  consists  of  21  terms  ;  the  sum  of  the  three 
terms  in  the  middle  is  129,  and  of  the  last  three  is  237  :  find 
the  series.  Ans.  3,  7,  11, 83. 

19.  The  first  term  of  an  A.  P.  is  5,  and  the  fifth  term  is 
11  :  find  the  sum  of  8  terms.  Ans.  82. 

20.  The  sum  of  four  terms  in  A.  P.  is  44,  and  the  last 
term  is  17  :  find  the  terms.  Ans.  5,  9,  13,  17. 

21.  The  seventh  term  of  an  A.  P.  is  12,  and  the  twelfth 
term  is  7 ;  the  sum  of  the  series  is  171  :  find  the  number  of 
terms.  Ans.  18,  or  19. 

22.  A  sets  out  from  a  place  and  travels  2^-  miles  an  hour. 
B  sets  out  3  hours  after  A,  and  travels  in  the  same  direction, 
3  miles  the  first  hour,  3J  miles  the  second,  4  miles  the  third, 
and  so  on.    In  how  many  hours  will  B  overtake  A?     Ans.  5. 

23.  In  the  series,  1,  3,  5,  etc.,  the  sum  of  2n  terms  :  the 
sum  of  n  terms  ;:  x  i  1;  find  the  value  of  x,  Ans.  4. 


336  EXAMPLES. 

24.  Find  an  A.  P.  such  that  the  sum  of  the  first  five  terms 
is  one-fourth  the  sum  of  the  following  five  terms,  the  first 
term  being  unity.  Ans.  1,  —  2,  —5, —26. 

25.  If  the  sum  of  m  terms  of  an  A.  P.  be  always  to  the 
sum  of  n  terms  in  the  ratio  of  m2  to  w2,  and  the  first  term  be 
unity,  find  the  ?ith  term.  Ans.  2?i  —  1. 

26.  If  2w  +  1  terms  of  the  series  1,  3,  5,  7,  9,  ....  be 
taken,  show  that  the  sum  of  the  alternate  terms,  1,  5,  9, 

will  be  to  the  sum  of  the  remaining  terms  3,  7,  11, 

as  n  -f-  1  to  n. 

27.  On  the  ground  are  placed  n  stones ;  the  distance 
between  the  first  and  second  is  one  yard,  between  the 
second  and  third  three  yards,  between  each  of  the  remain- 
ing stones  five  yards  :  how  far  will  a  person  have  to  travel 
who  shall  bring  them  one  by  one  to  a  basket  placed  at  the 
first  stone  ? 

Ans.  5  n?  —  17  n  -j-  16  yards. 

28.  Find  a  series  of  arithmetic  means  between  1  and  21, 
so  that  their  sum  has  to  the  sum  of  the  two  greatest  of  them 
the  ratio  of  11  to  4.      Ans.  9  means,  3,  5,  7, 19. 

29.  Find  the  number  of  arithmetic  means  between  1  and 
19  when  the  second  mean  is  to  the  last  as  1  to  6.      Ans.  17. 

30.  If  the  second  term  of  an  A.  P.  be  a  mean  proportional 
between  the  first  and  the  fourth,  show  that  the  sixth  term 
will  be  a  mean  proportional  between  the  fourth  and  the 
ninth. 

Find  the  last  term  of  each  of  the  following  geometric 
series : 

31.  2,  —6,   18,    to  8  terms.        Ans.  —4374. 

32.  2,      3,     4J to  6  terms.  -\\3-. 

33.  3,  -32,    33, to  2n  terms.  -32\ 

Sum  the  following  scries  : 

34.  1,  -},  a, to  12    terms.  Ans.  Jf}£. 

35.  9,  -6,  4, to    7    terms.  5|f. 


EXAMPLES.  337 

36.  2,  -4,  8, to    2/)  terras.    Ans.  f(l  -  22*). 

37.  \/2,  V^6,  3^2, to  12  terms.       364(^6  +  V^2). 

38.  Insert  3  geometric  means  between  486  and  6. 

Ans.    162,  54,  18. 

39.  Insert  4  geometric  means  between  J  and  128. 

Ans.  },  2,  8,  32. 

40.  Insert  3  geometric  means  between  1  and  256. 

Ans.  4,  16,  64. 

41.  Insert  4  geometric  means  between  3  and  —729. 

Ans.  -9,  27,  -81,  243. 
Sum  the  following  series  : 


"648" 
2 


42.  f,      i,       £, to  6  terms.  ^l?is. 

43.  1,  — J,       J, to  infinity. 

44.  6,  —2,       |, to  infinity.  4-£. 

45.  -J,      §,      24t? t°  infinity.  1. 

46.  |,-1,        |, to  infinity.  ff. 

47.  .9,  .03,  .001, to  infinity.  f  J. 

Find  the  value  of  the  following  repeating  decimals : 

48.  .4282828  .  .  .   Ans.  f|f.  I      50.    .16.  Ans.  £. 

49.  .28131313  .  .  .         T%^.  I      51.    .378.  §£. 

52.  The  sum  of  three  terms  in  G.  P.  is  63,  and  the 
difference  of  the  first  and  third  terms  is  45  :  find  the  terms. 

Ans.  3,  12,  48;  or  36,  -54,  81. 

Let  a,  or,  ar2  denote  the  numbers. 

53.  The  sum  of  the  first  four  terms  of  a  G.  P.  is  40.  and 
the  sum  of  the  first  eight  terms  is  3280 :  find  the  series. 

Ans.  1,  3,  9, 

54.  The  sum  of  three  terms  in  G.  P.  is  21.  and  the  sum 
of  their  squares  is  189  :  find  the  terms.  Ans.  3,  6,  12. 

55.  A  person  who  saved  every  year  half  as  much  again  as 
he  saved  the  previous  year  had  in  seven  years  saved  8102.95  : 
how  much  did  he  save  the  first  year?  Ans*  §3.20. 


338  PERMUTATIONS  AND    COMBINATIONS. 


CHAPTER    XVII. 

PERMUTATIONS    AND    COMBINATIONS  — BINOMIAL 
THEOREM. 

PERMUTATIONS    AND    COMBINATIONS. 

166.  Definitions.  —  The  different  orders  in  which  a 
number  of  things  can  be  arranged,  either  by  taking  some  or 
all  of  them,  are  called  their  Permutations. 

Thus,  the  permutations  of  the  letters  a,  6,  c,  taken  one  at 
a  time  are  three,  viz.,  a,  6,  c;  taken  two  a  time,  are  six, 
viz.,  a6,  6a,  ac,  ca,  be,  cb ;  and  taken  three  at  a  time,  are 
also  six,  viz., 

abc,  acb,  bca,  bac,  cab,  cba. 

The  Combinations  of  things  are  the  different  groups  or 
collections  which  can  be  made,  either  by  taking  a  part  or  all 
of  them,  without  reference  to  the  order  in  which  the  things 
are  placed. 

Thus,  the  combinations  of  the  letters  a,  6,  c,  taken  two  at 
a  time  are  three,  viz.,  a6,  ac,  be;  ab  and  6a,  though  differ- 
ent permutations,  form  the  same  combination,  both  consisting 
simply  of  a  and  6  grouped  together. 

It  appears  from  this  that  in  forming  combinations  we  are 
concerned  only  with  the  number  of  things  each  group  con- 
tains ;  while  in  forming  permutations  we  have  also  to  consider 
the  order  of  the  things  which  make  up  each  group  ;  thus  the 
above  six  permutations  of  the  letters  a,  6,  c,  taken  three  at 
a  time,  form  but  one  combination. 

167.  The  Number  of  Permutations.  —  To  find  the 
number  of  permutations  of  n  different  things,  taken  r  at  a 
time. 

Let  the  different  things  be  represented  by  n  letters,  a,  6, 


THE  NUMBER   OF  PERMUTATIONS.  339 

r, Set  a  aside  ;  write  down  the  other  n  —  \  letters 

in  a  line  ;  put  a  before  each  of  them  in  succession  ;  we  thus 
obtain  ab,  etc,  ad,  etc.,  or  n  —  1  permutations,  each  of  two 
letters  in  which  a  stands  first.  In  the  same  manner  there 
are  n  —  1  permutations,  each  of  two  letters  in  which  b 
stands  first.  Similarly  there  are  n  —  1  permutations, 
each  of  two  letters  in  which  c  stands  first ;  and  so  on  for  each 
of  the  other  letters  ;  and  as  there  are  n  of  them,  the  whole 
number  of  permutations  of  the  n  letters,  two  together,  is 
n(n  -  1). 

Again,  set  a  aside,  and  group  the  other  n  —  1  letters,  two 
and  two ;  as  has  just  been  shown,  there  are  (n  —  1)  (to  —  2) 
such  groups.  Put  a  before  each  of  them,  and  we  have 
(to  —  1)0*  —  2)  permutations,  each  of  three  letters  in  which 
a  stands  first.  Similarly  there  are  (n  —  1)(to  —  2)  per- 
mutations, each  of  three  letters  in  which  b  stands  first ;  and 
so  on  for  each  of  the  other  letters.  Therefore  the  whole 
number  of  permutations  of  n  letters  taken  three  at  a  time  is 
n(n  _  i)(„  _  2). 

Proceeding  thus,  and  noticing  that  at  any  stage,  the 
number  of  factors  is  the  same  as  the  number  of  letters  in 
each  permutation,  and  that  the  negative  number  in  the  last 
factor  is  one  less  than  the  number  of  letters  in  each  permu- 
tation, we  shall  have  the  number  of  permutations  of  n  things, 

r  together  equal  to  n(n  —  1)  (n  —  2) to  r  factors  ; 

and  the  rth  factor  is  n  —  (r  —1)  or  n  —  r  -f-  1- 

Hence,  the  whole  number  of  per  mutations  of  n  things  taken 
r  at  a  time  is 

n(n  _  i)(7l  _  2) (to  —  r  +  1)  .     .   (1) 

If  all  the  letters  are  taken  together,  r  =  to,  and  (1) 
becomes 

w(to  -  1)(to  -  2) 3-2-1   .     .     .   (2) 

Hence,  the  number  of  permutations  of  n  things  taken  all  at 
a  time  is  equal  to  the  product  of  the  natural  numbers  from  1 
up  to  n. 


340  THE   NUMBER    OF   COMBINATIONS. 

It  is  usual  to  denote  this  product  by  the  symbol  In,  which 
is  read  "  factorial  ?*."  *     Thus, 

Factorial  6,  or  [6,  means  6  •  5  •  4  •  3  •  2  •  1,  or  720  ; 
factorial  5,  or  15,  means  5  •  4  •  3  •  2  •  1,  or  120. 

From  the  law  of  formation  it  is  clear  that  [7  =  716. 

More  generally,  \n  -f  1  =  (n  -f-  1)[^« 

Thus  |  n  +  1  contains  all  the  factors  of  In,  and  one  factor, 
n  +  1,  additional. 

Denoting  the  number  of  permutations  of  n  things  taken  r 
at  a  time  by  the  symbol  nPr,  we  have  from  (1)  and  (2) 

BPr=n(n-l)(n-2) (n-r+1);     and    nPn=[n. 

Thus,  BP4=n(n— l)(n— 2)(»— 3). 

Also    »P6=nP4(n— 4}  =n(n- 1)  (n  — 2)  (n  — 3)  (n-4)  ; 

and  so  on. 

1.  Four  persons  enter  a  railway  carriage  in  which  there 
are  six  seats  ;  in  how  many  ways  can  they  take  their  places  ? 

Here  n  =  6,  and  r  =  4  ;  then  by  (1)  we  have 

6P4  =  6  •  5  •  4  •  3  =  3G0. 

2.  Required  the  number  of  changes  which  can  be  rung, 
(1)  with  5  bells  out  of  8,  and  (2)  with  the  whole  peal. 

Ans.  (1)  6720;   (2)  40320. 

3.  Required  the  number  of  different  ways  in  which  6 
persons  can  be  seated  at  a  dinner  table.  Ans.  720. 

168.  The  Number  of  Combinations.  —  To  find  the 
number  of  combinations  of  n  different  things  taken  r  at  a 
time. 

The  number  of  permutations  of  n  things  taken  rata  time 
is 

n(n  -  l)(n  —  2) (n  -  r  +  1).    (Art.  167). 

But  each  combination  of  r  things  taken  r  at  a  time  will 

*  It  is  also  sometimes  denoted  by  n !. 


THE   NUMBER   OF   COMBINATIONS.  341 

make  \r  permutations,  by  (2)  of  Art.  167;  therefore  there 
are  \r  times  as  many  permutations  as  combinations.  Hence, 
calling  mGr  the  required  number  of  combinations,  we  have 

n(n-  l)(n-2) (n-r+1) 

This  formula  for  nCr  may  also  be  written  in  a  different 
form  ;  for  if  we  multiply  the  numerator  and  the  denominator 
by  the  product  of  the  natural  numbers  from  1  up  to  n  —  r, 
it  becomes 

_  n(n-l)(n-2) (n-r+l)(n-r) 2-1 

[r.(n-r)(»-r-l) 2-1 

The  numerator  is  now  the  product  of  the  natural  numbers 
from  n  to  1,  or  is  In  (Art.  167);  the  denominator  is  the 
product  of  the  natural  numbers  from  /*  to  1,  and  from  n  —  r 
to  1.     Hence  we  have  , 

0  _         &         (2) 

»°'  -  [r\n-r  V  ' 

It  will  be  convenient  to  use  (1)  for  nCr  in  all  cases  where 
a  numerical  result  is  required,  and  (2)  when  it  is  sufficient 
to  leave  it  in  an  Algebraic  shape. 

Note  1.  —  If  in  (2)  we  put  r  —  n,  we  have 

\n  \ 

but  nCn  =  1,  so  that  if  the  formula  is  to  be  true  for  r  =  n,  the 
symbol  10  must  be  considered  as  equivalent  to  1. 

The  number  of  combinations  of  n  things  taken  r  at  a  time 
is  the  same  as  the  number  of  them  taken  n  —  r  at  a  time. 
For  the  number  taken  n  —  r  at  a  time  is,  from  (2), 

a       -  \n  -       ^  m 

n   n~r        {n  -  r  ,n  -  {n  -  r)        \n  -  r  \  r       '    K  ' 


which  =  RCr,  from  (2). 

The  truth  of  this  proposition  is  also  evident  from  the 
consideration  that  for  every  different  group  of  r  things  taken 
out  of  n  things  there  is  always  left  a  different  group  of 


342     TO   DIVIDE  111  -h  n    THINGS   INTO    TWO    CLASSES. 

n  —  r  things.  Hence  the  number  of  groups  of  r  things  out 
of  n  must  be  the  same  as  the  number  of  groups  of  n  —  r 
things.     Such  combinations  are  called  complementary. 

Note  2.  — Put  r  =  n;  then  from  (2)  and  (3),  nCn  =  nC0  =  1. 

The  proposition  just  proved  is  useful  in  enabling  us  to 
abridge  Arithmetic  work.     Thus, 

1.  Required  the  number  of  combinations  of  20  things 
taken  18  together. 

The  required  number  is  the  same  as  the  number  taken  2 

t0S°ther-  n         20  X  19        ,Qn 

If  we  had  used  the  formula  20O18,  we  should  have  had  to 
reduce  an  expression  whose  numerator  and  denominator 
each  contained  18  factors. 

2.  From  12  books,  in  how  many  ways  can  a  selection  of  5 
be  made  when  one  specified  book  is  always  excluded? 

Since  the  specified  book  is  always  to  be  excluded,  we  have 
to  select  the  5  books  out  of  the  remaining  11.     Hence 

=  11.10-9.8.7  =  462, 
11   5         1.2.3.4.5 

3.  How  many  combinations  may  be  made  of  10  letters 
taken  6  at  a  time?  Ans.  210. 

4.  From  11  books,  in  how  many  ways  can  a  selection  of 
4  be  made?  Ans.  330. 

169.  To  Divide  m  +  n  Things  into  Two  Classes. 

—  To  find  the  number  of  ivays  in  which  m  +  n  different 

things  can  be  divided  into  two  classes,  so  that  one  may  contain 

m  and  the  other  n  things. 

This  is  equivalent  to  finding  the  number  of  combinations 

of  m  -f  n  things  m  at  a  time,  for  every  time  we  select  one 

group  of  m  things,  we  leave  a  group  of  n  things  behind. 

Hence  by  (2)  of  Art.  168, 

\m  -h  n 
The  required  number  = 


[m  [n 


PERMUTATIONS  OF  n  THINGS  NOT  ALL  DIFFERENT.    343 

In  a  similar  manner  it  may  be  shown  that  the  number  of 
ways  in  which  m  +  n  +  p  different  things  can  be  divided 
into  three  classes  containing  m,  n,  p  things  respectively  is 

\m  4-  n  +  p 


\m  \n  \p 

1.  There  are  three  bookshelves  capable  of  containing  14, 
22,  and  24  books  ;  in  how  many  ways  can  60  books  be 
allotted  to  the  shelves? 

Here  we  have  to  divide  60  things  into  groups  of  14,  22, 
and  24  things. 

160 
Hence  the  required  number  =  = . 

1  \U  J 22  1 24 

2.  From  7  Englishmen  and  4  Americans  a  committee  of 
6  is  to  be  formed,  containing  2  Americans ;  in  how  many 
ways  can  this  be  done  ? 

Here  we  have  to  choose  2  Americans  out  of  4,  and  4 
Englishmen  out  of  7.  The  number  of  ways  in  which  the 
Americans  can  be  chosen  is  4<72 ;  and  the  number  of  ways  in 
which  the  Englishmen  can  be  chosen  is  7C4.  Each  of  the 
first  groups  can  be  associated  with  each  of  the  second. 
Hence  the  required  number  of  ways  is 

«0» x  ,Ci  =  fi  x  jig  =  j2]2|§ =  210, 

3.  In  how  many  ways  can  the  52  cards  in  a  pack  be 

152 
divided  among  4  players,  each  to  have  13?  Ans.  _  —  . 

°      F    J  [[13]4 

170.  Permutations  of  n  Things  not  all  Different. 

—  To  find  the  number  of  permutations  of  n  things  taken  all 
at  a  time,  when  they  are  not  all  different. 

Let  there  be  n  letters ;  and  suppose  p  of  them  to  be  a,  q 
of  them  to  be  b,  r  of  them  to  be  c,  and  the  rest  to  be 
unlike. 


344   PERMUTATIONS  OF  n  THINGS  NOT  ALL  DIFFERENT. 

Let  P  be  the  required  number  of  permutations.  If  in 
any  one  of  the  actual  permutations,  the  p  letters  a  were  all 
changed  into  p  letters  different  from  each  other  and  from 
all  the  rest,  then  from  this  single  permutation,  without  alter- 
ing the  position  of  any  of  the  remaining  letters,  we  could 
form  \p  new  permutations.  Hence  if  this  change  were  made 
in  each  of  the  P  permutations,  there  would  be  P  X  \p  per- 
mutations. 

Similarly,  if  in  any  one  of  these  new  permutations,  the  q 
letters  b  were  changed  into  q  letters  different  from  each  other 
and  from  all  the  rest,  then  from  this  single  permutation  we 
could  form  to  new  permutations.  Hence  the  whole  number 
of  permutations  would  now  be  P  X  \p  X  \q. 

In  like  manner,  if  the  r  letters  c  were  also  changed  so  that 
no  two  were  alike,  the  total  number  of  permutations  would 
be  P  X  \p  X  [g  X  |r.  But  this  number  must  be  equal  to 
the  number  of  permutations  of  n  different  things  taken  all 
together,  which  is  \n.     Hence 

Px[pX[gx|r  =  [». 

...   p-    L» 


\p\q\r_ 

And  similarly  any  other  case  may  be  treated. 

1.  How  many  different  permutations  can  be  formed  out  of 
the  letters  of  the  word  Mississippi  taken  all  together? 

Here  we  have  11  letters,  of  which  4  are  i,  4  are  s,  and  2 
are  p. 

|H 

•*•     P  =   [4  |4>   =  11 -10-9.  7-5  =  34650. 

2.  How  many  different  permutations  can  be  made  out  of 
the  letters  of  the  word  assassination  taken  all  together? 

Ans.  10810800. 

3.  How  many  different  permutations  can  be  made  out  of 
the  letters  of  the  word  Heliopolisf  Ans,  453600. 


BINOMIAL    TIIEOREM.  345 

BINOMIAL    THEOREM. 

171.  Positive  Integral  Exponent.  —  The  method  of 
raising  a  binomial  to  any  power  by  repeated  multiplication 
has  been  explained  in  Art.  104.  We  shall  now  prove  a 
formula  known  as  the  Binomial  Theorem,*  by  which  any 
binomial  can  be  raised  to  any  power  without  the  labor  of 
actual  multiplication. 

To  'prove  the  Binomial  Theorem  for  a  positive  integral 
exponent. 

By  actual  multiplication  we  obtain 
{x  +  a)  (x  +  b)  —  x2  +  (a  +  b)x+ab, 
(x+a)  (x+b)  (x+  c)  =  xs  +  (a  +  b  +  c)x2  +  (ab  +  ac  +  bc)x  +  abc. 

In  th^st,  r~  alfs  we  see  that  the  following  laws  hold : 

1.  The  number  of  terms  on  the  light  side  is  one  more  than 
the  number  of  the  binomial  factors  on  tJie  left  side. 

2.  The  exponent  of  x  in  the  first  term  is  the  same  as  the 
member  of  binomial  factors,  and  decreases  by  one  in  each 
successive  term. 

3.  The  coefficient  of  the  first  term  is  unity ;  of  the  second 
term,  the  sum  of  the  letters  a,  b,  c;  of  the  third  term,  the  sum 
of  the  products  of  the  letters  a,  b,  c,  taken  two  at  a  time;  and 
the  fourth  term  is  the  product  of  all  the  letters. 

We  shall  now  prove  that  these  laws  always  hold  whatever 
be  the  number  of  binomial  factors. 

Suppose  these  laws  to  hold  for  n  —  1  binomial  factors,  so 
that 

{x+a)  (x+b)  .  .  [x+Tc)  =xn~l+Axn-2+Bxn-*+Cxn-A+  .  .  K,  (1) 
where  A  —  a  +  b  +  c  +  ....+k,  the  sum  of  the  secoud 
-erms, 

B  =  ab  +  ac  +  be  + ,  the  sum  of  the  products 

of  these  terms  taken  two  at  a  time. 

C  —  abc  +  abd  + ,  the  sum  of  the  products  of 

these  terms  taken  three  at  a  time. 

K  =  abed fe,  the  product  of  all  these  terms. 

*This  theorem  was  discovered  by  Newtou. 


346  POSITIVE  INTEGRAL  EXPONENT. 

Multiply  both  sides  of  (1)  by  another  factor  (x  +  l);  thus, 

(x+a)(x+b) {x+k)(x+l)=x«+{A+l)x*-1 

+  (B+Al)xn-2+(C+Bl)xn-*+ +KI.  .  .  (2) 

Now  i  +  /  =  a  +  6  +  c+ +  k  +  1 

=  the  sum  of  all  the  terms  a,  b,  c, I. 

B  +  Al  =  ab+ac+bc+  .  .  .  +al+bl+cl+  .  .  .  +  kl 
.=  the  sum  of  the  products  taken  two  at  a  time. 

C  +  Bl  =  abc  +  abd  -f  .  ■  .  +  abl  +  acl  -f  bcl  +  .  .  ., 
=  the  sum  of  the  products  taken  three  at  a  time. 


Kl  =  abed  .  .  .  kl  =  the  product  of  all  the  terms 
a,  6,  c,  ....  I. 

Also  the  exponent  of  x  in  the  first  term  is  the  same  as  the 
number  of  binomial  factors,  and  decreases  by  1  in  each  suc- 
cessive term.    • 

Hence  if  the  laws  hold  when  n  —  1  factors  are  multiplied 
together,  they  hold  when  n  factors  are  multiplied  together ; 
but  they  have  been  proved  to  hold  for  3  factors,  therefore 
they  hold  for  4  factors,  and  therefore  for  5  factors,  and  so 
on,  generally,  for  any  number  whatever.* 

Now  let  6,  c,  d, ?,  each  =  a ;  then  the  binomial 

factors  are  all  equal,  and  the  first  member  of  (2)  becomes 

(x+a)  (x+a)  . . .  =  (x+a)  taken  n  times  as  a  factor  =  (x+a)n ; 

and  the  second  member  becomes 

A  +  l  =  a  +  a  +  a  +  ....  =  a  taken  n  times  =  na. 

B  +  Al  =  aa  +  aa  ■+• .  .  .  =  a2  taken  as  many  times  as  there 

are  combinations  of  n  letters  taken  2  at  a  time  =  n^n  ~  ' 
(Art.  181).  IA 

C  +  Bl  =  aaa  +  aaa  -f =  a3  taken  as  many 

times  as  there  are  combinations  of  n  letters  taken  3  at  a 

time  =  "(»-  \Hn-2)     and  go  Qn 

H 

*  Tins  method  of  proof  in  called  Mathematical  Induction. 


POSITIVE   INTEGRAL    EXPONENT.  347 

Kl  =  aaaa  .  •  •  •  =  a  taken  n  times  as  a  factor  =  an. 
Substituting  in  (2),  we  obtain 

(x  +  a)n  =  as"  -f  nax"-1  +  n0l  ~  *)  aV"* 

+  n(n-l)(n-2)fl^.<  +  B>>rfl    §         (8) 

E 

This  formula  is  called  the  Binomial  Tlieorem;  the  series 
in  the  second  member  is  called  the  expansion  of  (x  +  a)n. 
In  this  expansion  we  observe  the  following 

Rule. 

(1)  Tlie  exponent  of  x  in  the  first  term  is  the  same  as  the 
exponent  of  the  power,  and  decreases  by  unity  in  each  succeed- 
ing term;  the  exponent  of  a  begins  with  one  in  the  second 
term,  and  increases  by  unity  in  each  succeeding  term. 

(2)  The  coefficient  of  the  first  term  is  1,  that  of  the  second 
is  the  exponent  of  the  power,  and  if  the  coefficient  of  any 
term  be  multiplied  by  the  exponent  of  x  in  that  term,  and  the 
product  be  divided  by  the  number  of  the  term,  the  quotient 
will  be  the  coefficient  of  the  next  term. 

By  changing  x  to  a  and  a  to  x,  we  have 

(a  +  x)n  =  an  +  nan~lx  +  n(n  ~  ^  o"- V 

+  n(n-l)(n-2)fl„.a,  +  >>>af   §        (4) 

If  we  write  —a  for  a  in  (3),  we  obtain 
(x  -  a)n  =  af  -  nax"-1  +  n(n  =  ^aV-2  - (5) 

Thus  the  odd  powers  of  a  are  negative  and  the  even  powers 
positive,  and  the  last  term  is  positive  or  negative  according 
as  n  is  even  or  odd. 

Suppose  a  =  1,  then  (4)  becomes 
/i   i     \n     -I   .         .  n(n  —  l)   o  .  w(n  — l)(n  — 2)   •  ,'  -    //>>. 

|2  [3 

which  is  the  simplest  form  of  the  binomial  theorem. 


348  GENERAL    TERM   OF  THE  EXPANSION. 

1.  Expand  (x  +  a)5. 
By  the  rule,  we  have 

(x  +  a)5  =  x5  +  bxAa  +  10A2  +  lOara3  -f  5xa*  +  a6. 

Similarly 

2.  (a-2a;)7=a7-7a6(2a;)+21a5(2a;)2-35a4(2«)8+35a3(2x;: 

-21a2(2a;)5+7a(2a;)6-(2a;)7. 
=a7-14a6a;+84aV-280a4a;3+560a3a;4 

-672a2a;5+448aa;6-128a;7. 

Expand  the  following  by  the  Binomial  Theorem  : 

3.  (jc  _  3)5.     Ans.  x5-  15a;4-f  90a;3  -  270a;2  +  405a;  -  243. 

4.  (Sx+2y)K       81x*  +  2l6x3y  +  2Wx2y2  +  96xy*  +  16y4. 

5.  (a;2  +  a;)5.  a;10  -f  5a;9  +  10a;8  +  10a;7  +  oxG  -f  a;5. 

6.  (2 -fa;2)4.  16  -  48a;2  +  54a;4  -  27a;6  +  f£a;8. 

The  sum  of  the  coefficients  in  the  expansion  of  (1  +  x)*  is  2n.    For 
put  x  =  1 ;  then 

(1  +  x)»  =  (1  +  1)»  =  2»  =  1  +  n  +  ^LgliJ  +  etc. 

If 
ss  sum  of  the  coefficients. 

Also,  by  putting  x  =  —  1,  we  have 
(l-l)»  =  l-n+2iiL^l>-etc; 

.*.  0  =  sum  of  the  odd  coefficients  —  the  sum  of  the  even  ones; 
i.e.,  the  sums  of  the  odd  and  even  coefficients  are  equal,  and  therefore 
each  =  1x2"  =  2n~\ 

172.  The  rth  or  General  Term  of  the  Expansion. 

—  In  the  expansion  of  (a;  +  a)n,  we  see  that  the  second 

terra  is  nxn~1a  ;    the  third  terra  is  n^n  "~~ — 'xn~2a2;  and  so 

If 
on ;    the  last  factor  in  the  denominator  of  each  term  being 

one  less  than  the  number  of  the  term  to  which  it  applies,  one 
greater  than  the  negative  number  in  the  last  factor  of  the 
numerator,  and  the  same  as  the  exponent  of  a ;   and  also 


GENERAL    TERM    OE    THE   EXP  AS  SI  OS.  349 

that  the  exponent  of  x  is  found  by  subtracting  the  exponent 
of  a  from  n.     Hence  the 

n(n  -  l)(n-  2) (n  -  r  +  2)jfl-r  +  1ar~1 


»•'■'  term 


r  -  1 


This  is  called  the  general  term,  because  by  giving  to  r 
different  numerical  values,  any  assigned  term  may  be  ob- 
tained. 

Tlie  coefficient  of  the  r*h  term  from  the  beginning  is  equal  to 
the  coefficient  of  the  i*1  term  from  the  end. 

The  coefficient  of  the  ?*th  term  from  the  beginning  is 

n(n  —  l)(n  —  2) (n  -  r  -f  2) 

By  multiplying  both  terms  by  \n  —  r  4-  1,  this  becomes 

\n 
■ .     See  (1)  and  (2)  of  Art.  168. 

\r  -  1  \n  -  r  +  1 

The  ?,th  term  from  the  end  is  the  (n  -  r  +  2)th  term  from 
the  beginning,  and  its  coefficient  is 

n(n  —  1) r      .  .  .      ,  \n 

— '- ,  which  also  =  , ■ . 

\n  —  r  +  1  \r  -  1  \n  -  r  +  1 

Therefore  the  coefficients  of  the  latter  half  of  an  expansion 
may  be  taken  from  the  first  half. 

1.  Find  the  fifth  term  of  (a  -f  2x3)17. 
Here  n  —  17,  r  =  5  ;  therefore  the 

5th  term  =  17  * 16  ' 15  '  14oM  x  ltx*  =  38080a13x'12. 
1-2.3.4 

2.  Find  the  14th  term  of  (3  -  a)15.  Ans.  -945au. 

3.  Find  the    7th  term  of  (a3  +  3ab)9.  61236a156G. 

4.  Find  the    5th  term  of  (a2  -  &2)12.  495a1668. 

5.  Find  the    5th  term  of  (3x*  -  4^)9.  126  x  35u:-:  IV 


350  EXAMPLES. 

Rem.  —  In  the  demonstration  (Art.  171)  we  assumed  n  to  denote  a 
positive  integer.  But  the  Binomial  Theorem  is  also  true  when  n  is 
a  positive  fraction,  or  a  negative  quantity  whole  or  fractional.  For 
the  proof  of  the  Binomial  Theorem  for  fractional  or  negative  values 
of  n,  the  student  is  referred  to  the  College  Algebra  (Art.  192). 

EXAMPLES. 

1.  How  many  different  numbers  can  be  formed  by  using 
six  out  of  the  nine  digits,  1,  2,  3, 9?       Ans.  60480. 

2.  Required  the  number  of  changes  which  can  be  rung 
upon  12  bells  taken  all  together.  Ans.  479001600. 

3.  Required  the  number  of  combinations  of  24  different 
letters  taken  4  at  a  time.  Ans.  10626. 

4.  Out  of  14  men,  in  how  many  wa3*s  can  11  be  chosen? 

Ans.  364. 

5.  How  many  different  products  can  be  formed  with  any 
three  of  the  figures  1,  3,  5,  7,  9?  Ans.  10. 

6.  In  how  many  ways  can  6  copies  of  Horace,  4  of 
Virgil,  and  3  of  Homer  be  given  to  13  boys,  so  that  each 
boy  may  receive  a  book?  Ans.  60060. 

7.  Out  of  7  consonants  and  4  vowels,  how  many  words 
can  be  made  each  containing  3  consonants  and  2  vowels  ? 

Ans.  25200. 

8.  How  many  parties  of  12  men  each  can  be  formed 
from  a  company  of  60  men?  [60 

AnS'  [12  [48* 

9.  Out  of  12  Republicans  and  16  Democrats,  how  mauy 

different  committees  could  be  formed,  each  consisting  of  3 

Republicans  and  4  Democrats?  112  |16 

Ans.  ■  'TTn    x 


10.  Out  of  10  consonants  and  4  vowels,  how  many  words 
can  be  formed,  each  containing  3  consonants  and  2  vowels? 

Ans.  86400. 

11.  There  are  10  candidates  for  6  vacancies  in  a  com- 
mittee :  in  how  many  ways  can  a  person  vote  for  6  of  the 
candidates?  Ans,  210. 


EXAMPLES.  351 

12.  In  how  man)-  ways  can  a  cricket  eleven  be  chosen  out 
of  fourteen  players?  Ans.  364. 

13.  In  how  many  ways  could  2  ladies  and  2  gentlemen  be 
chosen  to  make  a  set  at  tennis  from  a  party  of  4  ladies  and 
0  gentlemen?  Ana.  90. 

14.  In  how  many  ways  could  2  ladies  and  2  gentlemen  be 
chosen  to  make  a  set  at  tenuis  from  a  party  of  G  ladies  and 
8  gentlemen?  Ana.  420. 

15.  From  6  ladies  and  5  gentlemen,  in  how  many  ways 
could  you  arrange  sides  for  a  game  of  croquet,  so  that  there 
would  be  2  ladies  and  one  gentleman  on  each  side  ? 

16  16 
Ans.  > ■  2  \aJ3  '  or  180°* 

16.  Out  of  G  ladies  and  8  gentlemen,  how  many  different 

parties  can  be  formed,  each  consisting  of  3  ladies  and  4 

gentlemen  ?  16  18 

Ans. 


ULLiL     m± 

17.  If  the  number  of  permutations  of  n  things  taken  4 
together  is  equal  to  12  times  the  number  of  permutations  of 
n  things  taken  2  together:  find  n.  Ans.  6. 

18.  In  how  many  ways  can  a  party  of  6  take  their  places 
at  a  round  table  ?  Ans.  60. 

19.  How  many  words  of  6  letters  may  be  formed  with  3 
vowels  and  3  consonants,  the  vowels  always  having  the  even 
places?  Ans.  36. 

Expand  the  following  by  the  Binomial  Theorem. 

20.  (2x  -  y)5. 

Ans.  32x-5  -  80x*y  +  80xY  -  40a,-y  +  lOxy*  -  if. 

21.  (3a  -|)6. 

Ans.  729a6  -  972a5  +  540a4  -  160a3  +  —  -  —  +  — . 

3  27        729 

22.  (1  +  2x  -  x2)\ 

Ans.  1  +8z+20z2+8£8-26:c4-8£6+20a;6-8.c7+a;8. 

23.  (3z2-2«z+3a2)3. 

Ans.  27a6-54az5+117a2z4-l  16aV+117a4a;2-54a6a:+27a6. 


352 


EXAMPLES. 


Expand  to  4  terms  : 

24.  (1  -  «)i 

25.  (1  -  3x)K 

26.  (1  -  3x)~K 

27-(i+iT- 

28.  (1  +  i«)-4. 

29.  (8  +  12a)5. 

30.  (9  -  6x)~K 

31.  (4a  -  8a)-i 

Write  down  and  simplify  : 

32.  The  4th  term  of  (x  - 

33.  The  10th  term  of  (1  - 

34.  The  4th  term  of 


Ans.  1  —  \x  —  -^x2  — 


T2T^ 


1 

—  X  —  ar 

1  ~  f  *3- 

1 

+ 

x  -f  2a;2 

+  tfa» 

1 

— 

x  +  fa;2 

-  i^ 

1 

— 

2a  -f  |as 

'■  —  fa8. 

4(1  +  a  -  |a2  +  J«3), 
*0  +  ®  +  K  +  ff  a;3), 
sb   ,   3     ^    ,    5     a;3 
2  "  a2  +  2 


JA+- 

2an        « 


a8/ 


5)13. 
-  2a;)12. 


(i + »y 


35.    The  7th  term  of 


'4x 


36.  The  5th  term  of  (xkr 

37.  The  8th  term  of  (1  -f 

38.  The  5th  term  of  (3a  - 

39.  The  14th  term  of  (210 


AY 

"  2a;  J  " 

ft  _  y*b-ty. 

ft 


2x) 

-  2b)~\ 


Ans.  -35750a;10. 
-112640a;9. 

40a763. 

1 0500 
a;3    " 

70x-y°«-2&-6. 

_  4  2  9  7.7 

16fr4 
243a5' 
-1848a;13. 


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